Recalling the idea of inverse relations
An inverse relation $R^{-1}$R−1 is a relation where all of the elements $x$x and $y$y in set of ordered pairs $\left(x,y\right)$(x,y) in the relation $R$R switch positions - the $y$y elements move into the first position and the $x$x elements move into the 2nd position of each ordered pair .
This means, for example, that the inverse of the relation $y=2x+3$y=2x+3 is given by $x=2y+3$x=2y+3. Ordered pairs of this relation such as $\left(1,5\right)$(1,5), $\left(-3,-3\right)$(−3,−3) and $\left(0,3\right)$(0,3), become, for the inverse relation, $\left(5,1\right)$(5,1), $\left(-3,-3\right)$(−3,−3) and $\left(3,0\right)$(3,0).
Note that the inverse relation $x=2y+3$x=2y+3 can be re-expressed with $y$y as the subject, so that $y=\frac{1}{2}\left(x-3\right)$y=12(x−3).
A relation and its inverse form mirror images of each other across the line $y=x$y=x.
Recall that a function is a relation where, for each $x$x value, there is assigned a unique $y$y value. Functions are either one-to-one relations (such as $y=2x+3$y=2x+3) or many-to-one relations (such as $y=x^2$y=x2).
This means that the inverse of a one-to-one function will retain the one-to one property, but the inverse of a many-to-one function will not. For example, the inverse of $y=2x+3$y=2x+3 is $y=\frac{1}{2}\left(x-3\right)$y=12(x−3), but the inverse of $y=x^2$y=x2 is $y=\pm\sqrt{x}$y=±√x.
Logs and Exponentials - mutual inverses
The inverse of the log function turns out to be an exponential function, and therefore the inverse of the exponential function becomes a log function. They are mutual inverses.
We will illustrate with some examples:
Example 1
The inverse of something like $y=3^x$y=3x is easy to determine from the definition of logarithms.
If $y=3^x$y=3x, then equivalently $x=\log_3y$x=log3y. Thus the inverse is given by $y=\log_3x$y=log3x.
Example 2
If $y=3^{x-2}-5$y=3x−2−5, then we can write $y+5=3^{x-2}$y+5=3x−2, and again from the definition, $x-2=\log_3\left(y+5\right)$x−2=log3(y+5).
Thus the inverse becomes
$y-2=\log_3\left(x+5\right)$y−2=log3(x+5)
$y=\log_3\left(x+5\right)+2$y=log3(x+5)+2
Example 3
Working in reverse, we can determine the inverse of the function like $y=\log_2\left(x-1\right)$y=log2(x−1) as follows.
In this example, note the clear and concise way the procedure is written :
| $R:$R: $y$y | $=$= | $\log_2\left(x-1\right)$log2(x−1) |
| $\therefore R^{-1}:$∴R−1: $x$x | $=$= | $\log_2\left(y-1\right)$log2(y−1) |
| $2^x$2x | $=$= | $y-1$y−1 |
| $y$y | $=$= | $2^x+1$2x+1 |
That is to say, the inverse of the log function $y=\log_2\left(x-1\right)$y=log2(x−1) is the exponential function $y=2^x+1$y=2x+1. We can go further and show the relationship with their graphs.
Note that the domain and range of the inverse function is the range and domain of the function respectively. The asymptotes are likewise swapped around.
Example 4
The inverse of the function $f\left(x\right)=2\log_e\left(x+1\right)-3$f(x)=2loge(x+1)−3 can be similarly found as follows. Before we start, it is best to write the function as $y=2\log_e\left(x+1\right)-3$y=2loge(x+1)−3:
| $R:$R: $y$y | $=$= | $2\log_e\left(x+1\right)-3$2loge(x+1)−3 |
| $\therefore R^{-1}:$∴R−1: $x$x | $=$= | $2\log_e\left(y+1\right)-3$2loge(y+1)−3 |
| $\frac{x+3}{2}$x+32 | $=$= | $\log_e\left(y+1\right)$loge(y+1) |
| $e^{\frac{x+3}{2}}$ex+32 | $=$= | $y+1$y+1 |
| $y$y | $=$= | $e^{\frac{x+3}{2}}-1$ex+32−1 |
| $f^{-1}\left(x\right)$f−1(x) | $=$= | $e^{\sqrt{x+3}}-1$e√x+3−1 |
Mutual inverses generalised
Starting with a general log function given by $y=\log_b\left(x-h\right)+k$y=logb(x−h)+k (omitting the dilation factor $a$a) we arrive at an expression for its generalised inverse as follows:
| $R:$R: $y$y | $=$= | $\log_b\left(x-h\right)+k$logb(x−h)+k |
| $\therefore R^{-1}:$∴R−1: $x$x | $=$= | $\log_b\left(y-h\right)+k$logb(y−h)+k |
| $x-k$x−k | $=$= | $\log_b\left(y-h\right)$logb(y−h) |
| $b^{x-k}$bx−k | $=$= | $y-h$y−h |
| $y$y | $=$= | $b^{x-k}+h$bx−k+h |
Had we started with $y=b^{x-k}+h$y=bx−k+h, we would have arrived at its inverse $y=\log_b\left(x-h\right)+k$y=logb(x−h)+k.
Applet
The applet below allows you to vary the three constants $b$b, $h$h and $k$k for the exponential function given by $y=b^{x-k}+h$y=bx−k+h and the log function given by $y=\log_b\left(x-h\right)+k$y=logb(x−h)+k.
The constants will automatically adjust each function's graph so you can see the various effects.
You should be able to discover that, depending on the values of these constants, a log function and its inverse (or alternatively an exponential function and its inverse) can either intersect in two places, become mutually tangential or else have no points in common at all.
For example, putting $h=k=0$h=k=0, you can find the approximate value of the base $b$b that makes $y=\log_bx$y=logbx and its inverse $y=b^x$y=bx mutually tangential. You should find that it is somewhere between $b=1.4$b=1.4 and $b=1.5$b=1.5. In fact, it can be proved that the exact base required is given by $e^{\frac{1}{e}}\approx1.444667861009763$e1e≈1.444667861009763.
Astoundingly, with $h=k=0$h=k=0, the point of tangency is $\left(e,e\right)$(e,e)!
Try other combinations of the constants to see what you can discover.
Worked Examples
Question 1
Consider the function $f\left(x\right)=2e^{x+3}$f(x)=2ex+3 for all $x\in\mathbb{R}$x∈ℝ.
By replacing $f\left(x\right)$f(x) with $y$y, find the inverse function.
Leave your answer in terms of $x$x and $y$y.
State the domain of the inverse function using interval notation.
State the range of the inverse function using interval notation.
Question 2
Rewrite $y=2\log_ex-3$y=2logex−3 with $x$x as the subject of the equation.
Question 3
Consider the function $f\left(x\right)=6\log_e5x-3$f(x)=6loge5x−3 for $x>0$x>0.
By replacing $f\left(x\right)$f(x) with $y$y, find the inverse function.
Leave your answer in terms of $x$x and $y$y.
State the domain of the inverse function using interval notation.
State the range of the inverse function using interval notation.