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Solve the equation below:

Mathematics
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\[\log_{4} x=1+\log_{2} 2x, x>0\]
i guess we must solve for x
@experimentX do u understand?

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yeah ... solve for x.
use this property of log \[ \log_a b = \frac{\log_kb}{\log_k a}\]
i still dont get it sorry
sorry ... error there \[ \log_4x = \frac{\log_2k}{\log_24} = \frac{\log_2k}{2} \]
Oops ... error again, that k is supposed to be x
if you know other few properties of log then it's should be easy for you!!
\[ \frac{\log_2x}{2} = \log_22 + \log_22x\] \[ \implies \log_2x = 2\log_22 + 2\log_22x \] \[ \implies \log_2x = \log_22^2 + \log_2(2x)^2 \] \[ \implies \log_2x = \log_2(2^2*(2x)^2) \] \[ \implies x = 2^2*(2x)^2 \]
\[ \implies x = \frac{1}{16} \]
yw ... though not sure about answer!! check on wolframalpha!!
its correct
because these are past papers they tell us the answer but they dont show us how to get it

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