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#### karush

##### Well-known member

- Jan 31, 2012

- 2,928

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