For what values are these logarithms true?

Question

For what values, x and y, are both these equations true?

$$\frac {\log(x)}{\log(y)} = \frac 23$$

AND

$$\frac xy = \frac 23$$

How would one solve this?

Answer 1

The first equation can be rewritten as $$\ln x = {2\over 3}\,\ln y$$ or $$x = e^{\ln x} = e^{2/3 \ln y} = \left( e^{\ln y}\right)^{2/3} = y^{2/3}.$$ Substituting this into the second equation yields $$y^{-1/3} = {y^{2/3} \over y} = {2\over3},$$ so $\displaystyle y^{1/3} = {3\over2}$ and $\displaystyle y=\left(3\over2\right)^3={27\over 8}$. Then $\displaystyle x={2\over3}\cdot y = {2\over3}\cdot{27\over 8} = {9\over 4}$.

Answer 2

Hint.
Notice that, if $x,y> 0$, then \begin{align*} \frac{\log x}{\log y} &= \frac{2}{3}\\ \log x &=\frac{2}{3}\log y\\ \log x &= \log \left(y^{2/3}\right) \end{align*} and $$x = \frac{2}{3}y.$$