+202 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
\(b=729.\)
+118667 \(log_b81=\frac{2}{3}\\ b^{log_b81}=b^{\frac{2}{3}}\\ 81=b^{\frac{2}{3}}\\ 9^2=b^{\frac{2}{3}}\\ \sqrt{9^2}=\sqrt{b^{\frac{2}{3}}}\\ 9=b^{\frac{1}{3}}\\ 9^3=b\\ b=729 \)
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\\
