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# simple question, but I can't figure it out

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

Apr 1, 2020

#1
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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
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$$log_b81=\frac{2}{3}\\ b^{log_b81}=b^{\frac{2}{3}}\\ 81=b^{\frac{2}{3}}\\ 9^2=b^{\frac{2}{3}}\\ \sqrt{9^2}=\sqrt{b^{\frac{2}{3}}}\\ 9=b^{\frac{1}{3}}\\ 9^3=b\\ b=729$$

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