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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

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  • $\begingroup$ Perhaps it is easier for you with $\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$. So $\frac{x}{\log(1/\sqrt{2})}=\log(2)$. $\endgroup$ Jan 18 '20 at 19:46
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Applying $\ln$, get $\ln({\sqrt2}/2)^x=\ln2\implies x\ln({\sqrt2}/2)=\ln2\implies x=\ln2/\ln({\sqrt2}/2)=1/\log_22^{-1/2}=1/(-1/2)=-2$.

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

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

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\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}

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

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