Question

given the function f(x)= 0.5(3)^x, what is the value of f^-1(7)

Answer 1

@thomaster

Answer 2

@campbell_st I opened a new one

Answer 3

would I have to do 0.5 times itself 3 times?

Answer 4

@mathmale

Answer 5

@kjones0331

Answer 6

@RadEn @SithsAndGiggles

Answer 7

\[f(x)=y=\frac{1}{2}(3)^x\]
Solve for \(x\) in terms of \(y\):
\[2y=3^x\\
\ln(2y)=\ln 3^x\\
\ln(2y)=x\ln 3\\
x=\frac{\ln(2y)}{\ln3}\\
x=\log_3(2y)~~\Rightarrow~~f^{-1}(x)=\log_3(2x)\]
Now plug in \(x=7\).

Or, more simply, just solve the following equation for \(x\):
\[7=\frac{1}{2}(3)^x\]

Answer 8

hmm ok so I subtract 7 from both sides

Answer 9

or divide 7 I believe

Answer 10

is it division @SithsAndGiggles

Answer 11

No, that's not how you would solve for \(x\). One of the first steps would be to take it down from the exponent:
\[7=\frac{1}{2}(3)^x\\
2\cdot7=2\cdot\frac{1}{2}(3)^x\\
14=3^x\]
To lower the exponent, you take the logarithm of both sides:
\[\ln14=\ln3^x\]
One of the properties of logarithms says \(\ln a^b=b\ln a\), which gives you
\[\ln14=x\ln3\]
Then,
\[\frac{\ln14}{\ln3}=\frac{x\ln3}{\ln3}\\
\frac{\ln14}{\ln3}=x\]
Another property of logarithms: \(\dfrac{\ln a}{\ln b}=\log_ba\), so,
\[x=\log_3(14)\approx\cdots\]

Answer 12

I got 4.67

Answer 13

No, that's not right...

Answer 14

2.402?

Answer 15

@SithsAndGiggles heloo

Answer 16

Yes, that's the right answer. Sorry