Question

Solve the equation below:

Answer 1

\[\log_{4} x=1+\log_{2} 2x, x>0\]

Answer 2

i guess we must solve for x

Answer 3

@experimentX do u understand?

Answer 4

yeah ... solve for x.

Answer 5

use this property of log
\[ \log_a b = \frac{\log_kb}{\log_k a}\]

Answer 6

i still dont get it sorry

Answer 7

sorry ... error there
\[ \log_4x = \frac{\log_2k}{\log_24} = \frac{\log_2k}{2} \]

Answer 8

Oops ... error again, that k is supposed to be x

Answer 9

if you know other few properties of log then it's should be easy for you!!

Answer 10

\[ \frac{\log_2x}{2} = \log_22 + \log_22x\]
\[ \implies \log_2x = 2\log_22 + 2\log_22x \]
\[ \implies \log_2x = \log_22^2 + \log_2(2x)^2 \]
\[ \implies \log_2x = \log_2(2^2*(2x)^2) \]
\[ \implies x = 2^2*(2x)^2 \]

Answer 11

\[ \implies x = \frac{1}{16} \]

Answer 12

@experimentX thanks

Answer 13

yw ... though not sure about answer!! check on wolframalpha!!

Answer 14

its correct

Answer 15

because these are past papers they tell us the answer but they dont show us how to get it