## anonymous one year ago What is the value of x in the equation.... (logarithms), equation will be posted in comments

1. anonymous

$\log_{8}x + \log_{4}x = 10/3$

2. kc_kennylau

Firstly, the law: $\Large\log_ab=\dfrac{\log_nb}{\log_na}$ Substitute $$\Large a=8$$, $$\Large b=x$$ to get: $\Large\log_8x=\dfrac{\log_nx}{\log_n8}$

3. kc_kennylau

Substitute $$\Large a = 4$$, $$\Large b = x$$ to get: $\Large\log_4x=\dfrac{\log_nx}{\log_n4}$

4. kc_kennylau

What would be a reasonable $$\Large n$$ to fit in?

5. anonymous

10?

6. kc_kennylau

$\Large \log_8x+\log_4x=\dfrac{10}3$ $\Large \dfrac{\log_nx}{\log_n8}+\dfrac{\log_nx}{\log_n4}=\dfrac{10}3$ I would actually choose $$\Large n = 2$$ because $$\Large 4=2^2$$ and $$\Large 8=2^3$$

7. kc_kennylau

Let's use $$\Large n = 10$$ anyway.

8. kc_kennylau

$\Large \dfrac{\log_{10}x}{\log_{10}2^3}+\dfrac{\log_{10}x}{\log_{10}2^2}=\dfrac{10}3$

9. anonymous

I was debating whether to use 2, but I didn't know how to use the 2 to the power of 2 and 3 in the equation.

10. kc_kennylau

okay, let's use $$\Large n = 2$$ then.

11. kc_kennylau

$\Large \dfrac{\log_2x}{\log_22^3}+\dfrac{\log_2x}{\log_22^2}=\dfrac{10}3$

12. kc_kennylau

$\Large \dfrac{\log_2x}3+\dfrac{\log_2x}2=\dfrac{10}3$

13. kc_kennylau

Are you able to find $$\Large \log_2x$$ now?

14. anonymous

Sorry, I'm not completely sure how to finish it.

15. anonymous

Would we simplify them into one fraction?

16. kc_kennylau

Let's substitute $$\Large a$$ as $$\Large \log_2x$$.

17. kc_kennylau

The equation then becomes: $\Large\frac a3+\frac a2=\frac{10}3$

18. anonymous

a=4?

19. kc_kennylau

Yes :)

20. anonymous

x=16!

21. kc_kennylau

So $$\Large \log_2x=4$$

22. kc_kennylau

Yes :D

23. kc_kennylau

Smart boy/girl

24. anonymous

Haha, girl - Thank you so much!

25. kc_kennylau

no problem :pp