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## GoldRush18 4 years ago Solve the equation below:

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1. GoldRush18

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

2. GoldRush18

i guess we must solve for x

3. GoldRush18

@experimentX do u understand?

4. experimentX

yeah ... solve for x.

5. experimentX

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

6. GoldRush18

i still dont get it sorry

7. experimentX

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

8. experimentX

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

9. experimentX

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

10. experimentX

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

11. experimentX

$\implies x = \frac{1}{16}$

12. GoldRush18

@experimentX thanks

13. experimentX

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

14. GoldRush18

its correct

15. GoldRush18

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

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