213k views
0 votes
If f(x) = log2 (x + 1), what is f-1(2)? (2 points)

User Jozcar
by
7.1k points

2 Answers

6 votes

Answer:

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

User Jack Guy
by
8.2k points
3 votes

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)

User Holmes Conan
by
7.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories