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GoldRush18
 4 years ago
Solve the equation below:
GoldRush18
 4 years ago
Solve the equation below:

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GoldRush18
 4 years ago
Best ResponseYou've already chosen the best response.0\[\log_{4} x=1+\log_{2} 2x, x>0\]

GoldRush18
 4 years ago
Best ResponseYou've already chosen the best response.0i guess we must solve for x

GoldRush18
 4 years ago
Best ResponseYou've already chosen the best response.0@experimentX do u understand?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0yeah ... solve for x.

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

GoldRush18
 4 years ago
Best ResponseYou've already chosen the best response.0i still dont get it sorry

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0sorry ... error there \[ \log_4x = \frac{\log_2k}{\log_24} = \frac{\log_2k}{2} \]

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0Oops ... error again, that k is supposed to be x

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0if you know other few properties of log then it's should be easy for you!!

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

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \implies x = \frac{1}{16} \]

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0yw ... though not sure about answer!! check on wolframalpha!!

GoldRush18
 4 years ago
Best ResponseYou've already chosen the best response.0because these are past papers they tell us the answer but they dont show us how to get it
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