Find the Equation of a Logarithm Function (y=klogx+c)

This chapter should be read in conjunction with the previous chapter on the logarithm function and its transformations.

the inverse of log

The logarithm function is a one-to-one function. In other words, if $y_1=\log x$y1​=logx and $y_2=\log x$y2​=logx then $y_1$y1​ and $y_2$y2​ are the same number. Thus, if we are given a value for $y$y, we can discover the value of $x$x that it came from without any ambiguity and we can say that for each $y$y, $x=f^{-1}\left(y\right)$x=f−1(y) where $f^{-1}$f−1 is the inverse function of the log function.

It must be the case that $f^{-1}\left(\log x\right)=x$f−1(logx)=x because whatever the log function did to $x$x is undone by the inverse function $f^{-1}$f−1. It is also true that $\log\left(f^{-1}\left(x\right)\right)=x$log(f−1(x))=x, for the same reason.

We know that the logarithm of a number to a base $b$b is just the power to which $b$b must be raised to give the number.

Thus, if $\log_b\left(f^{-1}\left(x\right)\right)=x$logb​(f−1(x))=x it follows by definition that $b^x=f^{-1}\left(x\right)$bx=f−1(x) and we have confirmed what we should have already suspected, that the inverse of the logarithm function $\log_bx$logb​x is the exponential function $b^x$bx.

When the graphs of these two functions are drawn it is seen that each is the reflection of the other in the line $y=x$y=x.

 

determining a log function from its graph

Given two points in the appropriate domain and range, there is only one logarithm function that the two points both belong to. Thus, if two points belonging to the function are known, then it is possible to derive the function in the form $y=k\log x+c$y=klogx+c.

It should be said, however, that the same function will have different parameters $k$k and $c$c depending on the chosen base.

 

Example 1

Suppose a logarithm function includes the points $(3,4)$(3,4) and $(5,4.5)$(5,4.5). If we choose the natural logarithm function $\ln$ln which has base $e\approx2.72$e≈2.72, what should the parameters $k$k and $c$c be in the function definition $f(x)=k\ln x+c$f(x)=klnx+c?

The given points allow us to write two valid equations.

$4=k\ln3+c$4=kln3+c
$4.5=k\ln5+c$4.5=kln5+c

Subtracting the first of these from the second, we eliminate $c$c and find $0.5=k(\ln5-\ln3)$0.5=k(ln5−ln3). Thus, $k=\frac{0.5}{\ln5-\ln3}\approx0.98$k=0.5ln5−ln3​≈0.98. Then, by substitution into the first equation, we have $4=0.98\ln3+c$4=0.98ln3+c and so, $c\approx2.92$c≈2.92.

The required function is $f(x)=0.98\ln x+2.92$f(x)=0.98lnx+2.92.

This should be checked by substituting the given $x$x-values into the formula.

 

Example 2

Suppose we wish to use $\log_2$log2​ to construct a function that contains the points $(0.5,8)$(0.5,8) and $(1,0)$(1,0).

Because the function value is $0$0 at $1$1, we know that the parameter $c$c is $0$0. We only need to determine the vertical dilation factor $k$k.

Now, $f(x)=k\log_2x$f(x)=klog2​x and therefore, $8=k\log_2\frac{1}{2}$8=klog2​12​. Since $\log_2\frac{1}{2}=-1$log2​12​=−1, we see that $k=-8$k=−8 and the function we are looking for is

$f(x)=-8\log_2x$f(x)=−8log2​x

Suppose now we decide to use the logarithm with base $\frac{1}{2}$12​ instead of $\log_2$log2​.

As before, we make the appropriate substitutions and obtain $8=k\log_{0.5}\frac{1}{2}$8=klog0.5​12​.

Since $\log_{0.5}\frac{1}{2}=1$log0.5​12​=1, we have $k=8$k=8 and the required function is

$f(x)=8\log_{0.5}x$f(x)=8log0.5​x

 

 

In general, you could, as an exercise, show that if

$g(x)=-k\log_bx$g(x)=−klogb​x

with $k$k a positive number, then it is also true that

$g(x)=k\log_{\frac{1}{b}}x$g(x)=klog1b​​x.

 

Worked Examples

Question 1

Question 2

Question 3