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

Question

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

Answer 1

Applying $\ln$, get $\ln({\sqrt2}/2)^x=\ln2\implies x\ln({\sqrt2}/2)=\ln2\implies x=\ln2/\ln({\sqrt2}/2)=1/\log_22^{-1/2}=1/(-1/2)=-2$.

Answer 2

Hint: $$ \frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}=2^{-\frac{1}{2}} $$

so you want $x$ such that: $$ \left(2^{-\frac{1}{2}}\right)^x=2^{-\frac{1}{2}x}=2 $$

Answer 3

\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}

Answer 4

Since $\sin45^\circ=\frac{1}{\sqrt{2}}=2^{-1/2}$, the answer is $\frac{1}{-1/2}=-2$.