NZ Level 7 (NZC) Level 2 (NCEA)
Find the Equation of a Logarithm Function
Lesson

Under certain circumstances, we are able to deduce a function's equation from a few known points the associated curve passes through. For example it may be possible when the form of the equation is known.

##### Example 1

Suppose we consider the curve of a function with the form $y=a\log_2\left(x\right)$y=alog2(x) which passes through the point $\left(16,12\right)$(16,12)

This point must satisfy the equation (that is, when substituted, will make the equation true). Thus we know that:

 $12$12 $=$= $a\log_216$alog2​16     $(1)$(1)

Since $\log_216=4$log216=4, we can say that $12=4a$12=4a and so $a=3$a=3. Thus the function becomes $y=3\log_2x$y=3log2x

The information we needed was a single point and the form of the equation. More complex forms of the log function requires more information.

In the next example, we need two pieces of information to uniquely identify the equation:

##### Example 2

Suppose we know that $\left(-12,0\right)$(12,0) and $\left(0,2\right)$(0,2) are points on the curve given by $y=\log_2\left(x+h\right)+k$y=log2(x+h)+k.

Recall that this is the translation of the basic curve $y=\log_2x$y=log2x horizontally and vertically a distance of $h$h and $k$k units.

Can we identify the equation.

If we substitute the points into the equation, we derive two equations as follows:

 $0$0 $=$= $\log_2\left(h-12\right)+k$log2​(h−12)+k     $(1)$(1) $2$2 $=$= $\log_2\left(h\right)+k$log2​(h)+k     $(2)$(2)

From equation $(1)$(1), we have, using the definition of a logarithm,  that $2^{-k}=h-12$2k=h12.

From equation $(2)$(2) we also have that $2^{2-k}=h$22k=h. Thus:

 $2^{2-k}$22−k $=$= $h$h $2^2\times2^{-k}$22×2−k $=$= $h$h $4\times\left[h-12\right]$4×[h−12] $=$= $h$h $4h-48$4h−48 $=$= $h$h $3h$3h $=$= $48$48 $h$h $=$= $16$16

Thus from $2^{-k}=h-12$2k=h12 we have $2^{-k}=4$2k=4 and this makes $k=-2$k=2

Hence the function's equation becomes $y=\log_2\left(x+16\right)-2$y=log2(x+16)2,  and the function is shown here:

##### Example 3

The function $y=a\log_bx$y=alogbx passes through the points $\left(16,12\right)$(16,12) and $\left(4,6\right)$(4,6). Is it possible to find the values of the constants $a$a and $b$b?

We can create two equations from the information given:

 $12$12 $=$= $a\log_b16$alogb​16     $(1)$(1) $6$6 $=$= $a\log_b4$alogb​4     $(2)$(2)

This looks like two pieces of distinct information, however upon closer inspection, we find that the two equations are really one and the same. Equation $(1)$(1) can be simplified:

 $12$12 $=$= $a\log_b16$alogb​16 $12$12 $=$= $a\log_b4^2$alogb​42 $12$12 $=$= $2a\log_b4$2alogb​4 $6$6 $=$= $a\log_b4$alogb​4

This is equation $(2)$(2).

Any other point chosen on this curve would lead to the same result. With one piece of information and two constants to find, we have a problem. We can only ever form a relationship between the constants - we simply cannot find unique values of the constants themselves:

For example both of our points $\left(16,12\right)$(16,12) and $\left(4,6\right)$(4,6) satisfy the two functions given by $y_1=3\log_2x$y1=3log2x and $y_2=6\log_4x$y2=6log4x. (Check this!)

So if $a=3$a=3 then $b=2$b=2 and if $a=6$a=6 then $b=4$b=4.

To explain this, we can show that the two  functions $y_1$y1 and $y_2$y2 are in fact exactly the same function.

Use this applet to check the above result.  Then try and find other pairs of log functions that are actually the same.

Using the change of base rule, we have, for $y_2$y2:

 $y_2$y2​ $=$= $6\log_4x$6log4​x $=$= $6\left(\frac{\log_2x}{\log_24}\right)$6(log2​xlog2​4​) $=$= $6\left(\frac{\log_2x}{2}\right)$6(log2​x2​) $=$= $3\log_2x$3log2​x $=$= $y_1$y1​

What all of this essentially means is that no matter how many known points we have for the curve $y=a\log_bx$y=alogbx we can never uniquely identify $a$a and $b$b.

#### Worked Examples

##### Question 1

Consider the function $F\left(x\right)=2^x$F(x)=2x.

1. Graph the function.

2. Add the line $y=x$y=x to the graph.

3. By reflecting points on the curve $F\left(x\right)=2^x$F(x)=2x over the line $y=x$y=x, draw the inverse of $F\left(x\right)=2^x$F(x)=2x.

4. What sort of graph have you drawn?

A

Reciprocal

B

Logarithmic

C

Exponential

D

A

Reciprocal

B

Logarithmic

C

Exponential

D
5. Hence state the equation of the logarithmic graph drawn.

##### Question 2

The graph of $f\left(x\right)=\log x$f(x)=logx (grey) and $g\left(x\right)$g(x) (black) is drawn below.

$g\left(x\right)$g(x) is a transformation of $f\left(x\right)$f(x).

1. What sort of transformation is $g\left(x\right)$g(x) of $f\left(x\right)$f(x)?

Horizontal translation.

A

Reflection.

B

Vertical translation.

C

Horizontal dilation.

D

Horizontal translation.

A

Reflection.

B

Vertical translation.

C

Horizontal dilation.

D
2. Hence state the equation of $g\left(x\right)$g(x).

##### Question 3

Consider the graph of $y=f\left(x\right)$y=f(x) drawn here.

1. When $x=1$x=1, what is the value of $y$y?

2. The answer to part (a) shows that the equation for $f\left(x\right)$f(x) is not of the form $\log_kx$logkx.

What transformation has been performed on the the graph of $y=\log_kx$y=logkx to get the graph of $f\left(x\right)$f(x)?

Vertical dilation

A

Horizontal translation

B

Vertical translation

C

Vertical dilation

A

Horizontal translation

B

Vertical translation

C
3. Hence, the equation of $f\left(x\right)$f(x) is of the form $f\left(x\right)=\log_kx+A$f(x)=logkx+A.

Find $A$A.

4. Determine the value of base $k$k in $f\left(x\right)=\log_kx+1$f(x)=logkx+1.

5. Hence, state the equation for $f\left(x\right)$f(x).

### Outcomes

#### M7-2

Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs

#### 91257

Apply graphical methods in solving problems