# [SOLVED]Inverse logarithm and find k

#### karush

##### Well-known member
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

Staff member
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

##### Well-known member
Re: inverse log and find k

well that was a better idea...