Question

If f(x) = log2 (x + 1), what is f-1(2)? (2 points)

Answer 1

3

Explanation:

`f^(-1)(2)=b \text{ implies } f(b)=2`

This we means to to solve the following equation for b:

`f(b)=\log_2(b+1)`

`2=\log_2(b+1)` since f(b)=2

Write in equivalent exponential form:

`2^2=b+1`

`4=b+1`

Subtract 1 on both sides:

`4-1=b`

`3=b`

This means
`f(3)=2 \text{ so } f^(-1)(2)=3`

You could actually find the inverse function if you want to then replace input for the inverse with 2.

`y=\log_2(x+1)`

Your logarithm has base 2 and input (x+1) and output y.

The equivalent exponential form is:

`2^(y)=x+1`

If we solve for x then at the end swap x and y we would have found the inverse function.

Let's do that:

`2^y=x+1</p><p>Subtract 1 on both sides:</p><p>[tex]2^y-1=x`

Swap x and y:

`2^x-1=y`

The inverse function of our given function is:

`f^(-1)(x)=2^x-1`

Now we need to replace x with 2:

`f^(-1)(2)=2^2-1`

`f^(-1)(2)=4-1`

`f^(-1)(2)=3`

Answer 2

as you already know, to get the inverse of any expression, we start off by doing a quick switcheroo on the variables, and then solve for "y".

`\bf \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^(log_a x)=x} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{f(x)}{y}=\log_2(x+1)\implies \stackrel{\textit{quick switcheroo}}{\underline{x}=\log_2(\underline{y}+1)}\implies 2^x=2^{\log_2({y}+1)} \\\\\\ 2^x=y+1\implies 2^x-1=\stackrel{f^(-1)(x)}{y} \\\\[-0.35em] ~\dotfill\\\\ 2^2-1=f^(-1)(2)\implies 3=f^(-1)(2)`