Question

What is the value of x in the equation.... (logarithms), equation will be posted in comments

Answer 1

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

Answer 2

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

Answer 3

Substitute \(\Large a = 4\), \(\Large b = x\) to get:
\[\Large\log_4x=\dfrac{\log_nx}{\log_n4}\]

Answer 4

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

Answer 5

10?

Answer 6

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

Answer 7

Let's use \(\Large n = 10\) anyway.

Answer 8

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

Answer 9

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.

Answer 10

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

Answer 11

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

Answer 12

\[\Large \dfrac{\log_2x}3+\dfrac{\log_2x}2=\dfrac{10}3\]

Answer 13

Are you able to find \(\Large \log_2x\) now?

Answer 14

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

Answer 15

Would we simplify them into one fraction?

Answer 16

Let's substitute \(\Large a \) as \(\Large \log_2x\).

Answer 17

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

Answer 18

a=4?

Answer 19

Yes :)

Answer 20

x=16!

Answer 21

So \(\Large \log_2x=4\)

Answer 22

Yes :D

Answer 23

Smart boy/girl

Answer 24

Haha, girl - Thank you so much!

Answer 25

no problem :pp