Given $x^2y = 32$ and $x^3/y = 1/8$, find $\log_2 x$ and $\log_4 y$

Question

Let $\log_2 x=a$ and $\log_4 y = b$. Then from

$$x^2y = 32$$

and

$$\frac{x^3}{y} = \frac{1}{8}$$

we need to find the values of $a$ and $b$.

I substituted values for $x$ and $y$:

$\log_2 2^2 = 2$ and $\log_4 8 = 1.5$ or $\log_2 8 = 2\cdot1.5$

where $a$ is $2$ and $b$ is $1.5$.

So $\log_2 x^2y=a+2b$ or $\log_2 32 = 2+(2\cdot1.5)$

If $\frac{x^3}{y} = \frac{1}{8}$ then the exponent would be $-3$, but this would mean $a = 2$ and $b = 1.5$ and would not prove to be true throughout the equation.

Any assistance is appreciated.

Answer 1

From

$$x^2y=32, \frac{x^3}{y}=\frac{1}{8}$$

we have

$$x^5 = 4$$

$$\Rightarrow x = 4^{\frac{1}{5}} = 2^{\frac{2}{5}}$$

Therefore, we have

$$ y = 8x^3 $$

$$= 8 \cdot (2^{\frac{2}{5}})^3 $$

$$= 2^3 \cdot 2^{\frac{6}{5}} $$

$$= 2^{\frac{21}{5}}$$

$$= 4^{\frac{21}{10}}$$

Hence, we have

$$\log_2(x) = \log_2 2^{\frac{2}{5}} = \frac{2}{5}$$

$$\log_4(y) = \log_4 4^{\frac{21}{10}} = \frac{21}{10}$$

Answer 2

More simply than taking logs, plugging $x=2^a$ and $y=4^b$ (as advised in comment) directly gives: $$32=2^{2a}4^b\iff2^5=2^{2a+2b}\iff5=2a+2b$$ and $$\frac18=2^{3a}4^{-b}\iff2^{-3}=2^{3a-2b}\iff-3=3a-2b.$$ The solution is therefore $$a=\frac25, \quad b=\frac{21}{10}.$$

Ted's suggestion (developped 20 min later in @ZYX's answer) may look even simpler, but is actually equivalent to solving this system the following way: $$\begin{align}(2a+2b=5,\quad3a-2b=-3)&\iff(5a=5-3,\quad2b=3a+3)\\&\iff\left(a=\frac25,\quad b=\frac{21}{10}\right).\end{align}$$