Name of School:
ROUTINE INFORMATION Inhlakanipho High
Student Surname and Name:
Maphumulo Mthokozisi
Student Number:
207503878
Grade:
10
Subject:
Mathematics
Topic:
Logs and logarithmic function
Content /Concept Area:
Exponents, logs and graphs.
CAPS page number:
57
Duration of Lesson:
45 minutes
Date Specific Aims: To develop skills of analysis of behaviour of a logarithmic function graph. To provide the opportunity to explore a logarithmic through its graph of y = log ax, where (a>0: a ≠1: x ¿ 0 ). To explore the relationship between the values of logarithmic expression to exponents To explore the shape of a graph and their function. Lesson Objectives:
Knowledge
Learners should gain
Skills Learners should be able to do
Value /Attitude Learners should acquire
knowledge and
following :
values and attitudes conducive to :
understanding of the following : Define, determine and analyse
Draw the graph of a function
the shape of the graph of logarithmic function Express logarithmic function into exponent form
Determine the value of the x within the logarithmic function
log ax Analyse exponents
the
function within
of the
logarithmic function: to draw them simultaneously o one set of system.
Acknowledge integrate
ability
to
logarithmic
expression to exponents.
Describe a function of a graph
Differentiate
logarithmic
Develop ability Draw, label
function
exponential
and
and
graphs
analyse
a
situation
involving both exponents and logs.
Approaches / Teaching Strategies:
Role plays. Case studies Class discussions. Questioning. Whole class-discussion Group Work Resources:
OHP/ Transparency Calculators Chalkboard Newspaper Worksheet Lesson Phases:
Introduction: To begin this lesson I will ask students to give examples of exponential graphs. I will ask open ended questions so that I give students a chance to express their response. The aim of asking these questions is to examine student’s background knowledge for exponents in relation to power. There is a basic rule for teaching logarithms which state that you cannot teach logarithm function without a prior understanding of exponents. So the aim of asking these questions is to let students revise exponents briefly. This lesson is done for preparation of logarithmic graphs which is the next lesson. You may skip a revision exercise below and jump to a topic of a development. Basically there are two main types of exponential function graph. These are i) ii)
exponential decay and exponential growth.
The graph of exponential decay is characterised by y = ax or f(x) = ax. In the exponential decay graph 0 ¿ a<¿ 1. Like all other function graphs it tend to have one to one relation, the input and output and a function given by f(x) = ax. In the graph of exponential decay where y = ax, the exponent x tend to acquire negative value e.g. y=2-x. The exponential decay graph on which y = b-x or y = ab-x or y= ab-x+p + q. For y=2-x, , remember that: 0 ¿ a<¿ 1 Y= 2-x , is an exponential decay and it is still a graph of f(x) = ax, where 0 ¿ a<¿ 1.
The above graph represents: - Y= 2-x, the exponential graph of an exponential decay. This is still a graph of f(x) = ax , 0 ¿ a<¿ 1 . The exponential decay graph above is an example of y=b -x , since y = 2-x and not y = ab-x. Please notice that you may ignore the value of a , where a is not provided but do always remember that the value of a takes the value of 1 (hint y = 2 -x ≡
y = 1x2-x, where 1
represents y – intercepts and also the y – component of a turning point) . Proof if y = 3x and y=2x3x are placed along the same graph, the value of a seem to reflect the value of ‘a’ to take a value of the y-intercept and turning point( hint for y = 2 x , a = 1 and turning point is ( x: y) or 0: 1) . The value of a has a significant difference in the graph. (Campell and Mcpetrie: 2012, pp 133) For any form of exponential graph we use exponential expression to denote the function of its graph. Most of books write the formula in full as y=abx+p + q. This formula often appear as y= bx. The values of an exponential function ab x+p + q, are always present, so if you do not see them consider the fact that they took the value of zero (0). NB sometimes the value of a behaves exceptional to y=abx+p + q. This is evident when one place the graph of y= b x and y= abx side by side. (You may ignore distinction between y= ab-x + q or y = abx + q).
Indeed an exponential graph function of y = 2
-x
has all values of abx+p + p. Students need to
take care of the following: 1) y= f(x) = 2-x. 2) y= abx+p + q, implies q = 0 and p= 0, if y = 2-x, ∴ the missing figures are always there. 3) NB: y=ax and y= abx and y= 2x are used interchangeable. 4) NB: y= a-x and y= ab-x and y= 2-x are also used interchangeable. 5) Where the equation of an exponential graph is written in form of either y=a x and y= abx or y= a-x and y= ab-x. You should have already known by now that : i) - ‘a’ stands for y – intercept, but also ‘a’ forms y-component of a turning point, if and only if y= abx stands for an exponential graph function. ***** ii) - that was exposed by those exponential graph function of y=4x2 x, y=3x2x, y= 2x when pulled together / drawn in one set of graph. A mysterious question is subject to interpretation for why does a takes a value of y-intercept together with the y-turning point whereas a function of y= bx implies (b>0, b
≠ 0). But one could reason that ‘b’ is not a nor ‘a’ is ‘b’.
In addition to that whether b = 3 or b =2 a graph does not take any form of b and make it yintercept or y-component of turning point. For example if, y = 3 x or y = 2-x: - the value of b does not take a value of y-intercept or turning point, instead you should get y-intercept = 1 and TP = (0:1) and this truly reflects that y = 3 x and y = 2-x resemble an exponential equation of a graph which seek to look like y= ab x where the value of a is ignored when the given function is y = bx. iii) - This delivers a message that exponential graph function and or equation of an exponential graph is paraplegic in nature. So this means that if y = bx you should be careful that b does not take the value of y-intercept instead it the value of ‘a’ is either hidden or ignored purposefully. Unfortunately it appear as if it becoming important when you are looking for a turning point. This is a mysterious discovery that has been left undiscovered in the entire evolution of mathematics. (We appeal to university professors to participate in consolidating this matter). The feedback from independent researchers regarding this matter would be given in the memorandum. But there is nothing wrong from what we have discussed above and this can boost students understanding of exponents and logarithmic. If you miss the understanding of something from logarithmic you often get struck from understanding logarithmic equations.
i)
exponential graph function rests on the rule that, for y= b x (where b> 0, b ≠ 1).
ii)
When dealing with graphs y= abx+p + q. know that you need to ignore the value of a when determining missing figures. However keep it in mind that a take a value of y-intercept and y-component of turning point. If the value of ‘a’ is absent then the value of a = 1. Consequently turning point is also (x: 1), which is likely to be (0: 1). For example y = 3 x , is same as y = a.3 x :- but ignore if your equation was not looking like y = 2.3 x. Bear in mind that y-intercept = a = 1.
iii)
Therefore y = 3x implies y= 1x3x. or y=1.3x or y=1(3x) = 3x.
iv)
Possible this hypothesis is supported by y = 3.2x.
The Hypothesis of Exponential Graphs : Since : y = 3.2x ≡
y = abx
a =3 is y-intercept and also y= y-component of a turning point (x: y) = (0: 3). When dealing with any exponential graph calculation you often ignore the value of ‘a’ if you were not given it, however bear in mind that you will need it when you express y- intercept or turning point. e.g. 1)
for : 1) y = abx y = 3x
≡
2) y = 3.2x
and 2) y= bx
y= bx , b = 3, a = 1 and TP is ( x: 1) or ( 0: 1) ≡
y = abx, a = 3 and TP is (x: 3) or ( 0: 3)
The gist point of this hypothesis is that in any exponential graph where y= b x, the value of ‘a’ should be 1 since y = 3.2x ≡ y = abx on which a take both the value y-intercept and turning point of exponential graph.
The application of the hypothesis of exponential function graph NB this rule applies if and only if this is an exponential graph of any type (decay or growth). This rule is applied where the value of q (otherwise known as asymptote) = 0. So if we want to apply it in
places where q change to become -2, -1, 1, 3 or any number greater than on- then we may need to adjust the values as follows. The value of a does not change but it is good to add a + q to determine the sum of these combination which must give us y-intercept and turning point. This theory confess that the value of ‘a’ is ‘a’ and not b or a +b. Therefore in a graph of y = a.b x + q, however the value of ‘a’ remain constant but the sum of a and q is what gives you y-intercept and turning point, mainly the y-component of a turning point. For y = a.bx+ q, we ignore the value of a but bearing in mind that the value of a act as the y-intercept of an exponential graph when q = 0. This means that a + q = y- intercept a point where a graph intersects (to cross) y- axis. At the same time the combination of a +q (the sum of a +q) become the y- component of a turning point or simple turning point (TP) of an exponential graph. . This implies that where y= 2-x -3, a = 1, so a + q = 1 + (-3) = -2. Therefore the actual value of a = -2 and the turning point is (0: -2). Therefore the value of ‘a’ of the exponential equation = the sum of a + q. The sum of a + q gives the y-intercept and turning point. To analyse the graph of y = 2x: with respect to y = a.bx + q. where q = 0. ( using the sum of a + q) Please do not forget to ignore the values of a where y = b x , you will only recall the value of a when you are looking for the combination of a and q for a turning point and y-intersect. In y = 2x , a = 1:- since y = 1.2x + 0. Combination is 1+ 0 = 1, therefore 1 is a y-intersection of an exponential graph to y-axis. At the same time, the combination of a+ q (1+ 0= 1) 1 is the turning point of an exponential graph. Therefore turning point is (x: y) = (0: 1). NB combination is not (2+ q) a = 1 and not b, b is something else which is the base of x ( bx). To analyse the graph of y = 2x2-x - 3: with respect to y = a.bx + q. where q is greater than or less than zero but not equal to zero, e.g q = -3 or q = 3. (Students cognitive development of their knowledge for exponential function). Leave this to students to do it as their classwork. Corrections of the above challenge goes as follows: Since y = a.bx + q, then a = 2 _ the combination we are looking for determination of y-intercept is a + q = 2+ (-3) = -1. This implies y-intercept = -1, meaning a graph intersects y –axis at y = -1. The component of y of the turning point of an exponential graph is also -1. Meaning TP is (0: -1).
The value of a is constant, meaning it cannot be changed, but when w make combination of a + q – we do that because we are looking to determine the turning point and y- intercept y= a.b x + q , the combination of the value of a = a+ q . If signs are not same you are encouraged to let them cancel each other – so take numbers as they are. The good news are that the combination of the sum of a + q still gives the value of y- intercept and also the value of the turning point. If q = zero (0) that fact may remain hidden. Why it is useful to master exponents when dealing with logarithmic expressions or logarithmic equations.
Importance of hypothesis Exponential Expression for the combination of a + q: This rule can change our thinking about how we perceive the drawing of exponential graph without reliance to input and output factors. It is also useful when determining the equation of an exponential graph. It is also a quickest method for classifying graph into exponential or decay. It can also reveal the effect of a in an exponential graph which is a part that mathematician seem to escape due to lack of the hypothesis discussed. NB students must be careful to use a multiplication sign when using ‘geo-gebra’ and not dots because they return erroneous figures. NB: the above rule applies if and only if the graph is exponential graph. This rule exposes exponential function graph or exponential equation to be paraplegic in nature. It also watch to reflect how some exponential graph value remain hidden from y= bx. y = abx +p + p: where these values are hidden you should accordingly assume they take the value of zero, except the value of a in the exponential graph. But the rule apply more or less similar in other graphs such as hyperbola, parabola and even quadratic equation.
So y = bx ≡
1) 2) 3)
y = abx ≡ y = 3x y= a.3x x ≡ y=3 y= 1.3x Where the value of ‘a’ represents:
y-intercept and y-value of a turning point TP( x: 1) for y = 3x, we acknowledge y =bx , ignore the value of ‘a’ but know that the value of a = 1. That one is both y-intercept and y-value of a turning point: - TP (0: 1). B is not a
4)
turning point and it is not y-intercept. Therefore y= 2.3x and y= 3x served us with beautiful point for discovering how, why and
5)
when does the value of a represents y-intercept. In this case y= abx and y = bx. This new rule tells us to ignore the value of a when dealing with y = bx, but also it tells us to bear in mind that for y = bx :- a = 1 , where a represents the y-component of turning
6)
point and also y-intercept. Y-intercept is a point where a graph intersects y-axis. This new rule also tells us that where y = abx, a =both the y-intercept and the ycomponent of the turning point. For y = 3.2x, a = 3 = y-intercept and also a y-component of the turning point (0: 3).
Where ‘b’ is concerned (hint look the formula of an exponential graph function) it takes the value of y- intercept and also forms the component of a turning point of an exponential function graph). These conditions take place only and only if this is an exponential function. NB sometimes we ignore the value of ‘a’ when y = b x and it appear as if this positions or window periods on which the value of ‘a’ is less than one. So this is thus far the most critical discovery we have found to be true. A grade 11 and 12 educator also agreed that sometimes a value is ignored and it can play two roles in the graph as explained above. Therefore it is always helpful to reflect the exponential function graph from its longest formula: y= abx+p + q. This revealed that exponential function graph is paraplegic in nature because it relies from other methods to derive its original function .Similarly hyperbolic function and parabolic functions were also found to be semi-permanent in nature. So they are also paraplegic. This need to be born in mind, when dealing with drawings of these graphs, full stop. This hypothesis has exposed that even hyperbole, parabola and quadratic graphs are paraplegic in nature. This is a hypothesis of paraplegic graphs system.
Development:Why did this hypothesis developed? This is due to the need to plot the exponential graph function without the use of a table method. We only rely on the table method during earliest high school grades such as grade 10. But as time goes by students learn to simple look at the value of ‘a’ and interprets what, why and how should a graph look like. The interpretations predates the form or shape of a graph mentally and graphically. There are two types of exponential graphs including exponential decay and exponential growth. The exponential graph of y = 2 -x represents exponential decay. We make distinction between values of ‘a’ and ‘b’ using the above hypothesis. Based on the theory of y=b x , a =1 or y= 1.2x). At the same time a is a y-component of the turning point meaning TP is (0:1). Then if ‘a’ is positive or a > 0 a graph mimic how a parabola is concave up .So this graph shall look up. Then for b is just ‘a’ base, where ‘a’ is a co-efficient of b x. That is where conversation of this lecture begins with our high school children. Exponential function involves exponential laws on which, y=ax
a
x
POWER Base
The exponential laws are as follows:
i)
If you want to simplify exponents, where bases have to be multiplied, see if bases are the same and add exponents. Therefore: am + an = am+ n,, the bases are the same, we add exponents.
ii)
If you want to simplify exponents, where bases have to be divided, see if bases are the same and subtract exponents. 2
Therefore:
b 1 b
= b2-1, the bases are the same we can equate
exponents. iii)
(am)n= (a)m n
iv)
a-1=
1 a
The application of exponents is evident in exponential decay and exponential growth. We have discussed the exponential decay to large extent. In this case, the exponential growth graph is given by y= 2x. Students are used to the system of the use of the input, the function and output. They must now to integrate their practical system to the applied method of the exponential graph as explained earlier from the above sections.
Exponential Decay:
x Y=2-x
-4 16
-3 8
-2 4
-1 2
0 1
1 0.5
2 3 0.2500 0.125 0
4 0.062 5
Y= 2-x, Exponential Decay. This is still a graph of f(x) = ax , (0 ¿ a<¿ 1) Basically the hypothesis of exponential graphs speak to both the exponential decay and exponential growth. The graph above is an exponential decay on which y= 2 -x. The assumption is that: For y= 2-x, ‘a’ is given by a process of imagination of y = a.b -x. So y = 1.2-x. therefore a = 1. In this exponential decay graph system a represents both the y-intercept and y –component of turning point. Therefore TP is (x: 1) or (0: 1) provided you have made your calculations. This method has its own proof as shown above.
Proof if y = 3x and y=2x3x are placed along the same graph, the value of ‘a’ seem to reflect the value of ‘a’ to take a value of the y-intercept and turning point (hint for y = 2 x, a = 1 and turning point is (x: y) or 0: 1). At the same time, it was observed that where y=2x3 x – the value of a remain to be a= 2, whereas a is the y-intercept of the exponential graph of y= 2x3-x and also the y-component of the turning point of y= 2x3 -x which is ( x:2) or (0: 2) The value of a has a significant difference in the graph. (Campell and Mcpetrie: 2012, pp 133). MN Maphumulo (2018) ‘PGCE-University of Natal : Edgewood Campus 2018’) Conclusion This hypothesis of exponential graphs was developed to enable the drawing of exponential graphs of decay and growth independent of application of input and output factors, where a
function of a graph is known or unknown. This theory adds that exponential graph is paraplegic in nature. It pulls example from trigonometric expressions of a function of y = sin
θ , y=cos θ
and y = tan θ
. This
become evident when one looks at eh values of special angles provided the radii are of the different size including r= 1 and r= 2. These values seem to take one but different value. The rationalization of the denominator of the special angles of trigonometric expression such as y = sin 45 seem to demonstrate that two values can appear in one common place from different specializations. For example, Sin 45 where r= 1 is same as sin 45 where r= 2. The rationalization of the real equations of trigonometric expressions of sin 45 gives equal values. This hypothesis also asserts that for y = bx, most values remain hidden. It say the value of ‘a’ is ignored when it takes the value of 1. However it must be born in mind that that value is required for determination of y-intercept and turning point. In this case this theory expand to point out that even hyperbola, parabola and quadratic formulae have common problem of being paraplegic. Then it is good to know the whole formula before you make calculations since the values of hidden values take the sum of (0). Therefore according to this hypothesis for y = 2-x, a = 1, but the entire formula is y = a.b x +p +q. Then if p and p are not given then their values are p= 0 and q = 0. These values are likely to assist in determining the asymptotes, axis of symmetry. Indeed the complexity behind the understanding of the axis of symmetry of exponential graph is now resolved. This is because it tend to have many response some of which seem to reflect how one understand hyperbola, parabola and even straight line and asymptotes. For many years this mystery has been hiding to be covered with respect to understanding of domain and range of a graph, full stop. Classwork: Exponential graph of exponential growth:x Y=2x
-4 -3 -2 -1 0.0625 0.1250 0.2500 0.500
0 1
1 2
2 4
3 8
4 16
Y= 2x, Exponential Growth y=ax, for a ¿ 0
Y= 2x, Exponential Growth y=bx, for a ¿ 0 In the given graph students are required to discuss why the exponential growth graph of y = 2x has the value of a = 1. Planned questions: i) ii) iii) iv) v)
What is the value of ‘a’ from the above graph? Use y = 2x, y = 3x2x and y = 4x2x to describe the effect of ‘a’ in an exponential graph. Identify whether these are exponential decay or exponential growth – exponential graph functions. What would you do to the graphs above to look concave down: y = 2x or y = 2-x . Take either of these graphs and illustrate. What are exponents and what are logarithmic equations?
Assessment: Homework, Classwork and Feedback is used to ass’s students as follows. Section A : student’s task. Solve for the following exponential graph equations and where requested draw graphs (it up to a teacher what he requesting: sometimes he can ask students to look for a value of a and use it to draw the y-intercept and a turning point). 1) a) y = 2x b) y = 2x + 1 c) y = 2. 2x – 2 2) In the following graph turn the value of a of each exponential graph equation so that all exponential graphs are looking upside down. a) y = 3-x
b) y = 2-x c) What is a difference between y = 2-x and y = 2x and how do these equations combined differ from differ from y = -2-x and y = -2x. (Hint: the effect of a, shape of a graph and type of exponential graph: decay vs growth). 3) a) Explain why there are diverse ways of expressing axis of symmetry in exponential graph using y = 3x across four different quadrants to support your answer. Section B: teachers task (feedback). More advanced questions: The two set of equations given below have equation of f(x) = bx + q. There is graph 1 and graph 2. In graph number 1 below, given the horizontal asymptote is x-axis, on which y = 0 i) ii) iii) iv) v)
Determine the value of ‘b’ Determine the co-ordinates of B Determine the equation of the exponential graph function given f (x) = y = bx + q. Without relying; express with reasons what is the value of a from the graph of y = bx . How would you advise us for the use of the formula of exponential graph above (encounter common errors of a and q in particular.
In graph number 2 below, given horizontal asymptote is y = 2, where y = bx + q is the equation of the exponential graph function. i) ii) iii) iv) v)
State down the value of q (q =?) (Hint: - refer to y = abx + q). What does basic rule for b apply from grade 10, 11 and 12 Caps book Determine the value of ‘b’ paying attention to the basic rule of b from a textbook. Determine the co-ordinates of ‘B’ Write down the equation of B
Section C: student’s task: Determine the equations of the exponential graphs below independent of teacher’s assistance: using grid reference as your guideline. Given y = 0 is horizontal asymptote.
Determine the equation of the graph below: The asymptote is y = 2 as indicated
Determine the equation of the graph below:
Section D: Challenge: is not part of a homework, need to be discussed before the class. From the hypothesis of exponential graph, given y = a.bx + q , where q = 0 , a takes the values of y-intercept and turning point. But if q becomes either 1, 2, 3…, .., …, To infinite value: you shall always add the value of a to a given value of q. Without the use of calculator or a graph predict what is y-intercept for y = 2. 3-x – 2 and also y = 2x + 1 Solution: is as follows: For y = a.bx+ q, we ignore the value of a but bearing in mind that the value of a act as the yintercept of an exponential graph when q = 0. This means that a + q = y- intercept a point where a graph intersects (to cross) y- axis. At the same time the combination of a +q (the sum of a +q) become the y- component of a turning point or simple turning point (TP) of an exponential graph. . This implies that where y= 2-x - 3, a = 1, so a + q = 1 + (-3) = -2. Therefore the actual value of a = -2 and the turning point is (0: -2). Therefore the value of ‘a’ of the exponential equation = the sum of a + q. The sum of a + q gives the y-intercept and turning point.
The end Memorandum: Planned Questions: Answers: i)
In an exponential graph of y = 2x , a = 1, because this graph is characterised by y = a.bx + q. a is characterised by 1.2 x + 0 …. a = 1 and not 2 because b stands in position of b and not a.
ii)
If the value of a increases from 1 to 4, where the value of q = 0, the value of ‘a’ becomes the y-intercept and also a turning point. In other hand the graph shape accelerate faster to come closer to the y-axis when the value of ‘a’ increases. The graph qualifies to be called y = a.bx + q because all values of ‘a’ endorse the fact that: a+ q = y- intercept and also the turning point.
iii)
These graphs are all exponential growth, since y = b x (where b>0, b
≠
1). The
x increases and y also increases. But also if the exponent of the base of b x is negative ( e.g. : ( b-x), we would get exponential decay , whereas when the exponent of the base of the function of exponential function is positive ( b +x) we would get an exponential graph. This is just a distinction between y = b-x or y = b+x but still the values of a remain constant. iv)
The value of a must be changed to be less than 1 (hint let a <1). Therefore y = -2 x or y = -2-x . This does not deviate from y = a.b x+p + q. , since a = -1, b = 2 and q = 0. Please notice -1 was not applied to ‘a’ and not to ‘b’ directly. The effect of a value of a where a < 0 implies that the graph is concave down.
v)
Exponents have base and power e.g.: ZF, F = power and Z = base. All bases have their co-efficient. in a.bx , a is a co-efficient of bx. For similar bases we add exponents: e.g.: Zd x Zq = Zd+q . (Only if you have to x Z xZ) For similar bases we may subtract exponents’ e.g.: Z d ÷ Zq , ( only if you have to divide z by z). ZqxFq can be written as (ZF) q
Logs and Logarithmic equations: Take for example the function on which y= a x and ignore everything that you know what we have said about a and b previously. This is a in a base position and not in a position of a coefficient of bx, such as y = a.bx. For y = bx rule implies that ( b>0, b ≠ 1). Likewise if any letter replace b to be y = b x this rule will apply similar to how it is understood above. So from y = ax the rule is applied as (a>0, b ≠ 1). The inverse of y = ax is ay = x . You simple take y and throw it in the exponent and vice versa. Likewise for y = bx , the inverse of this function would be x =by. Notice that b – the base of the powers did not change positions. The letter x chased y out of its position and leave a space to be taken by y. It is unclear how mathematician called this the change of the subject of a formula.
So the inverse for function of exponents and logarithmic function goes as follows:-
The inverse of y= bx is x= by . The base ‘b’ a power remain to its position. The inverse of y = ax is x=ay . The base ‘a’ remain constant to its position. This chasing change is defined as logarithmic function. In y=bx , where inverse is x = by the change is explained as log b x = y In y = ax, where inverse is x = ay the change is explained as log ax = y It clear that log entails change due to change made to enforce x to take the subject of a formula in the expression of a power where y = ax or y = bx. The log takes the constant base and equate it to term of a power. Therefore logax is bringing x and y back into their previous power bond, yet at the same time it shall solve for y.
x=ay ≡ logax = y. ( a is a constant base , and log always goes with base , carrying its former base to be equated to the power found on top of the base) So in x = ay , we could notice that y initially come across the equal sign whether LHS or RHS. So you log a base and carry a term across the equal sign (or any sign across the
equation) and equalize to the term on top of a base just in time/ that is found currently. Basically you always equalise with the term or a number found on top of a base just in time and not previously. So: x = ay x = by
≡ ≡
logax = y. logbx = y
This behaves like a CD radio which accommodate mp3 music and reject DVD indicating no DVD programmer. Know the difference between base, logarithm power and number. In y = bx, where x=by, where apply logbx = y For x=by:
i)
base (b) goes with a log to be log b, x appear across the equal sign and it is now going to be carried : log a x While x is getting carried position was changed so that previous previously bounded exponents are pulled together with x leaving it position getting pulled by the logarithmic function to be carried above the base of a log. The exponent y jumps off to its former positon and we are saying it is equated to the whole logarithmic expression. The rule does not inform students that y is always getting equated because it is an exponent, but the rule is subject to what stands in position of a power. If this was x a power, you would hence equated to x. So you must read these rules carefully.
ii) iii) iv) v)
Exponents: The exponents and logarithms refuse to lose their base – they have the same base. 5
32
2x
Log2x
Definitions: Exponents: For a > 0 and a ≠ 1as well as x > 0, y = logax means x= ay.
32
5
Note: it is better to move from power to exponent form than moving from exponent form to power. Students are strongly urged to adopt a style of expressing exponents before logarithmic expressions to pave their way. For y = loga x = y, a =base, x â&#x20AC;&#x201C; carried, y = change. so a y= x : instead of x= ay. . Hence it would be difficult to adopt other forms of expressions for students who are not orientated to function definition. Beside that it is confusing because it has no underlying factor of why is y the subject of a formula. But if you know what is base, you also know change from base and term being carried. Therefore:i)
the inverse of y = ax is x= ay, which is can be expressed as logarithmic form to logax=y . ( it is better to adopt style of adopting understanding of base , term being carried and change compared to y = logax).
ii)
The inverse of y = logax is x = logay , which could be expressed as a x =y . NB it is easy to move logarithmic form to exponential form by understanding of base, term being carried and change compared to y = ax.
Conclusion : if log55 = 1 : 5 1 = 5 , this is true for all values of log 5 equal to 1. But I cannot hide that this offer poor insight if an applicant does not expand with understanding of what does moving from logarithm to power entails in terms of base , carried term and change . For example: logaa = 1 : a1 = a Similarly : log24 = x : 2 x = 4 therefore 2x = 22 , the bases are the same , so x= 2 , we may equate exponents. Hence one can evaluate log28 by letting log28 = n â&#x20AC;Ś the solution is 2n = 8, n= 3 Hence evaluate log2 32. SS: hint let â&#x20AC;&#x201C; the log 2 32 = n, so 4n = 32, therefore n = 5/2 ( the bases are the same , we can equate the exponents: on the serious notice of power of 2) leave this challenge to students. Assessment: Section A: