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:
If we have an exponential of the form:
$b^x=a$bx=a
Then we can rewrite it as the logarithm:
$\log_ba=x$logba=x
and $b$b is called the base, just like in exponentials
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).
An important part to remember is that you can only take the logarithm of a positive number with a positive base that's not one:
$\log_ba$logba only makes sense when $a>0$a>0 and $b>0$b>0, $b\ne1$b≠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:
$\log_636=2$log636=2
Rewrite in exponential form: $\log_432=2.5$log432=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
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,
$\log_927=\frac{3}{2}$log927=32
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,
$\log_8\frac{1}{512}=-3$log81512=−3
Evaluate $\log_216$log216.
Rewrite the equation $\left(0.9\right)^0=1$(0.9)0=1 in logarithmic form.
Rewrite the equation $9^{\frac{3}{2}}=27$932=27 in logarithmic form.
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,
Exponential
$x$x | $y=2^x+1$y=2x+1 |
---|---|
$0$0 | $2$2 |
$1$1 | $3$3 |
$2$2 | $5$5 |
$3$3 | $9$9 |
Logarithmic
$x$x | $y=\log_2\left(x-1\right)$y=log2(x−1) |
---|---|
$2$2 | $0$0 |
$3$3 | $1$1 |
$5$5 | $2$2 |
$9$9 | $3$3 |
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.
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
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 |
Consider the function $f\left(x\right)=\log_5\left(x-3\right)$f(x)=log5(x−3) for all $x>3$x>3.
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.
Rewrite $y=2\log_ex-3$y=2logex−3 with $x$x as the subject of the equation.
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.