Question

log(sinxcosx)sinx*log(sinxcosx)cosx=1/4 sinxcosx is the base.

Answer 1

\[\large \log_{\sin x\cos x}(\sin x)\log_{\sin x\cos x}(\cos x)=\frac14\sin x \cos x\]just wanted to type it more clearly so I can think about it as I wake up...

Answer 2

After 1/4 there's no sinxcosx.

Answer 3

\[\large\log_{\sin x\cos x}(\sin x)\log_{\sin x\cos x}(\cos x)=\frac14\]good thing I typed it then :)

Answer 4

Sure is. :)

Answer 5

\[\frac{1}{\log_{\sin x}\sin x + \log_{\sin x}\cos x} \times\frac{1}{\log_{\cos x}\sin x + \log_{\cos x}\cos x} \]
\[\frac{1}{1 + \log_{\sin x}{\cos x}} \times \frac{1}{1 + \frac 1 {\log_{\sin x}\cos x}}\]
\[\log_{\sin x}\cos x = t\]
\[4t = (1 + t)^2\]

Answer 6

\[(t - 1)^2 = 0\]
\[t = 1\]
\[\log_{\sin x}\cos x = 1\]
\[\cos x = \sin x\]
\[x = \frac{\pi}{4}\]You can get other solutions as well, but I am lazy :/

Answer 7

The other solution is \[x=\frac{5\pi}{4}\]

Answer 8

Thank you!

Answer 9

You're Welcome!