## GoldRush18 2 years ago Solve the equation below:

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

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