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Find the base \(b\) so that \(log_b 81 = \frac{2}{3}.\)

 Apr 1, 2020
 #1
avatar+210 
+1

My knowledge of logarithms is a bit rusty but I think you can say that

\(b^{2/3}=81\).

Now to solve for \(b\) we need to take the square root of both sides and then cube it, giving you 

\(b^{2/3\cdot3/2}=81^{3/2}\)

The exponent of b will be just 1, so now we have

\(b=729.\)

.
 Apr 1, 2020
 #2
avatar+109525 
+1

\(log_b81=\frac{2}{3}\\ b^{log_b81}=b^{\frac{2}{3}}\\ 81=b^{\frac{2}{3}}\\ 9^2=b^{\frac{2}{3}}\\ \sqrt{9^2}=\sqrt{b^{\frac{2}{3}}}\\ 9=b^{\frac{1}{3}}\\ 9^3=b\\ b=729 \)

 

 

Which is exactly the same as Impasta got.    Thanks Impasta.

 

 

 

coding:

log_b81=\frac{2}{3}\\
b^{log_b81}=b^{\frac{2}{3}}\\
81=b^{\frac{2}{3}}\\
9^2=b^{\frac{2}{3}}\\
\sqrt{9^2}=\sqrt{b^{\frac{2}{3}}}\\
9=b^{\frac{1}{3}}\\
9^3=b\\

 Apr 3, 2020

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