How Do You Solve for k in a Logarithmic Function with a Given Inverse?

[karush]

Let \(\displaystyle f(x) = k\ log_2 x\)

(a) Given that \(\displaystyle f^{-1}(1)=8\), find the value of \(\displaystyle k\)

to get \(\displaystyle f^{-1}(x)\) exchange \(\displaystyle x\) and \(\displaystyle y\)

\(\displaystyle x=log_2 y^k\)

then convert to exponential form

\(\displaystyle 2^x=y^k \) then \(\displaystyle 2^{\frac{x}{k}} = y\)

so for \(\displaystyle f^{-1}(1) = 2^{\frac{1}{k}}= 8=2^3\) then \(\displaystyle \frac{1}{k}=3\) so \(\displaystyle k=\frac{1}{3}\)

(b) find \(\displaystyle f^{-1}\bigg(\frac{2}{3}\bigg)=2^{\frac{2}{3}\frac{3}{1}}=2^2=4\)

 

[MarkFL]

Re: inverse log and find k

a) Another approach would be to use that:

\(\displaystyle f^{-1}(1)=8\implies f(8)=1\)

and so:

\(\displaystyle f(8)=f\left(2^3 \right)=k\log_2\left(2^3 \right)=3k=1\,\therefore\,k=\frac{1}{3}\)

b) We could write:

\(\displaystyle f^{-1}\left(\frac{2}{3} \right)=x\)

\(\displaystyle f(x)=\frac{2}{3}\)

\(\displaystyle \frac{1}{3}\log_2(x)=\frac{2}{3}\)

\(\displaystyle \log_2(x)=2\)

\(\displaystyle x=2^2=4\)

Hence:

\(\displaystyle f^{-1}\left(\frac{2}{3} \right)=4\)

 

[karush]

Re: inverse log and find k

well that was a better idea...

 

[MarkFL]

Re: inverse log and find k

well that was a better idea...

I wouldn't say better, just different. :D

 

[CellCoree]

(c) The inverse of a logarithmic function is an exponential function. In this case, the inverse of f(x) is given by f^{-1}(x)=2^{\frac{x}{k}}. This means that for any input value x, the output of the inverse function will be the value of the exponent that, when raised to the base 2, will give the input value x. In other words, the inverse function "undoes" the action of the original function.

To find the value of k, we can use the fact that f^{-1}(1)=8. Plugging this into the inverse function, we get 2^{\frac{1}{k}}=8. Solving for k, we get k=\frac{1}{3}. This means that the inverse function is f^{-1}(x)=2^{\frac{x}{3}}.

To find the inverse of a specific input value, we can plug it into our inverse function. For example, to find f^{-1}\bigg(\frac{2}{3}\bigg), we plug \frac{2}{3} into our inverse function and get 2^{\frac{2}{3}\frac{3}{1}}=2^2=4. So the inverse of \frac{2}{3} is 4.

In conclusion, the inverse logarithm function is an exponential function, and the value of k can be found by setting f^{-1}(1) equal to a known value and solving for k. The inverse function can then be used to find the inverse of any input value.