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GoldRush18Best ResponseYou've already chosen the best response.0
\[\log_{4} x=1+\log_{2} 2x, x>0\]
 one year ago

GoldRush18Best ResponseYou've already chosen the best response.0
i guess we must solve for x
 one year ago

GoldRush18Best ResponseYou've already chosen the best response.0
@experimentX do u understand?
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
yeah ... solve for x.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
use this property of log \[ \log_a b = \frac{\log_kb}{\log_k a}\]
 one year ago

GoldRush18Best ResponseYou've already chosen the best response.0
i still dont get it sorry
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
sorry ... error there \[ \log_4x = \frac{\log_2k}{\log_24} = \frac{\log_2k}{2} \]
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
Oops ... error again, that k is supposed to be x
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
if you know other few properties of log then it's should be easy for you!!
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
\[ \frac{\log_2x}{2} = \log_22 + \log_22x\] \[ \implies \log_2x = 2\log_22 + 2\log_22x \] \[ \implies \log_2x = \log_22^2 + \log_2(2x)^2 \] \[ \implies \log_2x = \log_2(2^2*(2x)^2) \] \[ \implies x = 2^2*(2x)^2 \]
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
\[ \implies x = \frac{1}{16} \]
 one year ago

GoldRush18Best ResponseYou've already chosen the best response.0
@experimentX thanks
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
yw ... though not sure about answer!! check on wolframalpha!!
 one year ago

GoldRush18Best ResponseYou've already chosen the best response.0
because these are past papers they tell us the answer but they dont show us how to get it
 one year ago
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