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[SOLVED] Inverse logarithm and find k

karush

Well-known member
Jan 31, 2012
2,724
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Jan 31, 2012
2,724
Re: inverse log and find k

well that was a better idea...:cool:
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775