To find the inverse of the function [tex]\( f(x) = \frac{8^x}{2 + 8^x} \)[/tex], we proceed through the following steps:
### 1. Express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]
We start by setting [tex]\( y = f(x) \)[/tex]:
[tex]\[ y = \frac{8^x}{2 + 8^x} \][/tex]
### 2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]
We need to isolate [tex]\( x \)[/tex]. Start by manipulating the equation to move [tex]\( 8^x \)[/tex]:
[tex]\[
y = \frac{8^x}{2 + 8^x}
\][/tex]
Multiply both sides by [tex]\(2 + 8^x\)[/tex]:
[tex]\[
y(2 + 8^x) = 8^x
\][/tex]
Distribute [tex]\( y \)[/tex]:
[tex]\[
2y + y \cdot 8^x = 8^x
\][/tex]
Rearrange to get terms involving [tex]\( 8^x \)[/tex] on one side:
[tex]\[
2y = 8^x - y \cdot 8^x
\][/tex]
Factor [tex]\( 8^x \)[/tex] out on the right-hand side:
[tex]\[
2y = 8^x(1 - y)
\][/tex]
Solve for [tex]\( 8^x \)[/tex]:
[tex]\[
8^x = \frac{2y}{1 - y}
\][/tex]
### 3. Take the logarithm of both sides
To solve for [tex]\( x \)[/tex], take the natural logarithm on both sides:
[tex]\[
\ln(8^x) = \ln\left(\frac{2y}{1 - y}\right)
\][/tex]
Use the property of logarithms [tex]\( \ln(a^b) = b \cdot \ln(a) \)[/tex]:
[tex]\[
x \ln(8) = \ln\left(\frac{2y}{1 - y}\right)
\][/tex]
Isolate [tex]\( x \)[/tex]:
[tex]\[
x = \frac{\ln\left(\frac{2y}{1 - y}\right)}{\ln(8)}
\][/tex]
Since [tex]\( \ln(8) = \ln(2^3) = 3\ln(2) \)[/tex], this can be simplified to:
[tex]\[
x = \frac{\ln\left(\frac{2y}{1 - y}\right)}{3\ln(2)}
\][/tex]
### 4. Simplify the equation
Using the inverse functions and properties of logarithms, we need to evaluate the logarithm inside the fraction. The result consists of complex terms when considering all roots, providing a complete set of solutions:
[tex]\[
x = \left[ \frac{\ln\left(\left(-\frac{y}{y-1}\right)^{1/3}\left(-1 - \sqrt{3}i\right)\right) - \frac{2}{3}\ln 2}{\ln 2}, \frac{\ln\left(\left(-\frac{y}{y-1}\right)^{1/3}\left(-1 + \sqrt{3}i\right)\right) - \frac{2}{3}\ln 2}{\ln 2}, \frac{\ln\left(\frac{-2y}{y-1}\right)}{3\ln 2} \right]
\][/tex]
### Final Answer
[tex]\[
x = \left[ \frac{\ln\left(\left(-\frac{y}{y-1}\right)^{1/3}\left(-1 - \sqrt{3}i\right)\right) - \frac{2}{3}\ln 2}{\ln 2}, \frac{\ln\left(\left(-\frac{y}{y-1}\right)^{1/3}\left(-1 + \sqrt{3}i\right)\right) - \frac{2}{3}\ln 2}{\ln 2}, \frac{\ln\left(\frac{-2y}{y-1}\right)}{3\ln 2} \right]
\][/tex]
These three expressions together represent the complete inverse of the function [tex]\( f(x) = \frac{8^x}{2 + 8^x} \)[/tex], taking into account the different forms of the logarithmic solutions.
