6.03 Inverse relationship between exponentials and logarithms

Convert equations between log and exponential forms

We have learned about exponents (indices) and ways we can manipulate them in math. In math, we have pairs of inverse functions, or opposite operations, i.e., addition & subtraction, multiplication & division. The inverse function of an exponential is called a logarithm, and it works like this:

In other words, we use logarithms when we are interested in finding out the exponent needed ($x$x) to raise a certain base ($b$b) to a certain number ($a$a).

 

Worked examples

question 1

Express $6^2=36$62=36 in logarithmic form

Think: Remember, just as in all inverse operations, we want the exponent on its own and the logarithm on the other side.

Do:

$6$6 is the base and $2$2 is the exponent, so:

$\log_636=2$log6​36=2

question 2

Rewrite in exponential form: $\log_432=2.5$log4​32=2.5

Think: Remember that in exponential form we want isolate the resulting number after having the base raised to the exponent.

Do: Identify the components: $base=4$base=4 and the result of the logarithm is $exponent=2.5$exponent=2.5 . So,

$4^{2.5}=32$42.5=32

question 3

Rewrite the equation $9^{\frac{3}{2}}=27$932​=27 in logarithmic form.

Think: Remember, we want the exponent on its own and the logarithm on the other side.

Do: Identify the components: $base=9$base=9 and the exponent turns into the result $exponent=\frac{3}{2}$exponent=32​ . So,

$\log_927=\frac{3}{2}$log9​27=32​

question 4

Rewrite the equation $8^{-3}=\frac{1}{512}$8−3=1512​ in logarithmic form

Think: Remember, we want the exponent on its own and the logarithm on the other side.

Do: Identify the components: $base=8$base=8 and the exponent turns into the result $exponent=-3$exponent=−3 . So,

$\log_8\frac{1}{512}=-3$log8​1512​=−3

Practice questions

question 5

Question 6

question 7

 

 

Inverse relations

Recall that a relation and its inverse form mirror images of each other across the line $y=x$y=x. 

The inverse of the logarithms function $y=\log_2\left(x-1\right)$y=log2​(x−1) is the exponential function $y=2^x+1$y=2x+1. Note that the domain and range of the inverse function is the range and domain of the function respectively. The asymptotes are likewise reflections across the line $y=x$y=x.

When viewing the $x$x and $y$y tables for the logarithmic and exponential functions,

 we notice that the ordered pairs switch positions - the $y$y elements move into the first position and the $x$x elements move into the second position of each ordered pair.

 

Worked examples

question 8

What is the inverse of $y=3^{x-2}-5$y=3x−2−5?

Think: Before finding the inverse, we must isolate what we are to un-do - exponent. After we isolate the exponent, we need to identify the components of the equation.  

Do: We rewrite $y+5=3^{x-2}$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

question 9

Find the inverse of $f\left(x\right)=2\log_e\left(x+1\right)-3$f(x)=2loge​(x+1)−3 .

Think: Finding the inverse of a natural logarithmic function can be similarly found as the previous examples. Before we start, it is best to isolate the operation we want to un-do: natural logarithm.

Do: 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

 

Practice questions

Question 10

Question 11

Question 12