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

[karush]

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

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

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

$$x=log_2 y^k$$

then convert to exponential form

$$2^x=y^k $$ then $$2^{\frac{x}{k}} = y$$

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

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

 

[MarkFL]

Re: inverse log and find k

a) Another approach would be to use that:

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

and so:

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

b) We could write:

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

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

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

$$\log_2(x)=2$$

$$x=2^2=4$$

Hence:

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

 

[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