Use the properties of logarithms to write the expression as a single logarithm.

8 In d-2 Inx

hope it helps

Step-by-step explanation:

f−1(8)f^{-1}(8)f−1(8)f, start superscript, minus, 1, end superscript, left parenthesis, 8, right parenthesis.

To find f−1(8)f^{-1}(8)f−1(8)f, start superscript, minus, 1, end superscript, left parenthesis, 8, right parenthesis, we need to find the input of that corresponds to an output of 8888. This is because if f−1(8)=xf^{-1}(8)=xf−1(8)=xf, start superscript, minus, 1, end superscript, left parenthesis, 8, right parenthesis, equals, x, then by definition of inverses, f(x)=8f(x)=8f(x)=8f, left parenthesis, x, right parenthesis, equals, 8.

f(x)=3x+28=3x+2Let f(x)=86=3xSubtract 2 from both sides2=xDivide both sides by 3\begin{aligned} f(x) &= 3 x+2 8 &= 3 x+2 &&\small{\gray{\text{Let f(x)=8}}} 6&=3x &&\small{\gray{\text{Subtract 2 from both sides}}} 2&=x &&\small{\gray{\text{Divide both sides by 3}}} \end{aligned}f(x)862​=3x+2=3x+2=3x=x​​Let f(x)=8Subtract 2 from both sidesDivide both sides by 3​

So f(2)=8f(2)=8f(2)=8f, left parenthesis, 2, right parenthesis, equals, 8 which means that f−1(8)=2f^{-1}(8)=2f−1(8)=2f, start superscript, minus, 1, end superscript, left parenthesis, 8, right parenthesis, equals, 2

Finding inverse functions

We can generalize what we did above to find f−1(y)f^{-1}(y)f−1(y)f, start superscript, minus, 1, end superscript, left parenthesis, y, right parenthesis for any . [Why did we use y here?]

To find f−1(y)f^{-1}(y)f−1(y)f, start superscript, minus, 1, end superscript, left parenthesis, y, right parenthesis, we can find the input of that corresponds to an output of . This is because if f−1(y)=xf^{-1}(y)=xf−1(y)=xf, start superscript, minus, 1, end superscript, left parenthesis, y, right parenthesis, equals, x then by definition of inverses, f(x)=yf(x)=yf(x)=yf, left parenthesis, x, right parenthesis, equals, y.

f(x)=3x+2y=3x+2Let f(x)=yy−2=3xSubtract 2 from both sidesy−23=xDivide both sides by 3\begin{aligned} f(x) &= 3 x+2 y &= 3 x+2 &&\small{\gray{\text{Let f(x)=y}}} y-2&=3x &&\small{\gray{\text{Subtract 2 from both sides}}} \dfrac{y-2}{3}&=x &&\small{\gray{\text{Divide both sides by 3}}} \end{aligned}f(x)yy−23y−2​​=3x+2=3x+2=3x=x​​Let f(x)=ySubtract 2 from both sidesDivide both sides by 3​

So f−1(y)=y−23f^{-1}(y)=\dfrac{y-2}{3}f−1(y)=3y−2​f, start superscript, minus, 1, end superscript, left parenthesis, y, right parenthesis, equals, start fraction, y, minus, 2, divided by, 3, end fraction.