# Logarithms: How to solve for $x$ in $(\sqrt{2}/2)^x = 2$

Need some help on how to approach problem. My logs are rusty.

$$\log_{\sin45^\circ}2$$

I know that $$\sin 45° = \frac{\sqrt{2}}{2}$$. So then the equation becomes $$(\frac{\sqrt{2}}{2})^x = 2$$. How do I solve for $$x$$ in this case?

A step by step would be very helpful, thanks

• Perhaps it is easier for you with $\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$. So $\frac{x}{\log(1/\sqrt{2})}=\log(2)$. Jan 18 '20 at 19:46

Applying $$\ln$$, get $$\ln({\sqrt2}/2)^x=\ln2\implies x\ln({\sqrt2}/2)=\ln2\implies x=\ln2/\ln({\sqrt2}/2)=1/\log_22^{-1/2}=1/(-1/2)=-2$$.
Hint: $$\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}=2^{-\frac{1}{2}}$$
so you want $$x$$ such that: $$\left(2^{-\frac{1}{2}}\right)^x=2^{-\frac{1}{2}x}=2$$
\begin{align} & \frac {\sqrt 2} 2 = \frac 1 {\sqrt 2} \\[15pt] & \left( \frac 1 {\sqrt 2} \right)^2 = \frac 1 2 \\[15pt] & \left( \frac 1 {\sqrt 2} \right)^{-2} = 2 \end{align}
Since $$\sin45^\circ=\frac{1}{\sqrt{2}}=2^{-1/2}$$, the answer is $$\frac{1}{-1/2}=-2$$.