Find the Equation of a Logarithm Function

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 $\log_216=4$log2​16=4, we can say that $12=4a$12=4a and so $a=3$a=3. Thus the function becomes $y=3\log_2x$y=3log2​x. 

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=log2​x 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$2−k=h−12.

From equation $(2)$(2) we also have that $2^{2-k}=h$22−k=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$2−k=h−12 we have $2^{-k}=4$2−k=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=alogb​x 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​=3log2​x and $y_2=6\log_4x$y2​=6log4​x. (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=alogb​x we can never uniquely identify $a$a and $b$b.     

 

Worked Examples

Question 1

Question 2

Question 3