Logarithms

$\begin{aligned} \log_x \frac 18 & = - \frac 32 \\ \frac {\log_2 \frac 18}{\log_2 x} & = - \frac 32 \\ \frac {-3}{\log_2 x} & = - \frac 32 \\ \log_2 x & = 2 \\ \implies x & = 2^2 = \boxed{4} \end{aligned}$

Note first that $\large \log_{x} \left(\dfrac{1}{8}\right) = \log_{x} (8^{-1}) = -\log_{x}(8) = -\log_{x}(2^{3}) = -3\log_{x}(2)$ .

We then have that

$\large \log_{x} \left(\dfrac{1}{8}\right) = -\dfrac{3}{2} \Longrightarrow -3\log_{x}(2) = -\dfrac{3}{2} \Longrightarrow \log_{x}(2) = \dfrac{1}{2} \Longrightarrow x^{1/2} = 2 \Longrightarrow x = \boxed{4}$ .

Solution:

$\begin{aligned}{\log_{x}\frac{1}{8}}&=-\frac{3}{2} \\ \log_{x}1-\log_{x}8 &=-\frac{3}{2} \hspace{1cm} \color{#D61F06}{(\log_x\frac{a}{b}=\log_{x}{a}-\log_{x}{b})} \\ 0-\log_{x}8&=-\frac{3}{2}\hspace{1cm} \color{#D61F06}{(\log_{x}1=0)} \\ \log_{x}8&=\frac{3}{2} \\ 2\log_{x}8&=3 \\ \log_{x}8^{2}&=3 \hspace{1cm} \color{#D61F06}{(b\log_{x}a=\log_{x}a^{b})}\\ \log_{x}64&=3 \\ x&=\boxed{4} \hspace{1cm}\color{#D61F06}(log_{x}a=b \color{#D61F06}\implies{x^{b}=a})\end{aligned}$

Rewrite to: (√x)-3 = 2-3. Simplify to root x = 2 Answer: x = 4