Question

How do you add logs with different bases?

e.g. log (base 2) y + log (base 4) y = 6
Thanks in advance!

Answer 1

\[\frac{\log (y) }{\log(2)} + \frac{\log(y)}{\log4} = 6\]

Answer 2

:D Thanks.

Answer 3

i dont know how to go from here. lol

Answer 4

To be honest, I don't either. :P

Answer 5

lol, ok.. let me fig this out.

Answer 6

:D That's nice of you, thanks!

Answer 7

GOt it!
just a sec.

Answer 8

is it:
\[\large \log_2y + \log_4y = 6\]
?

Answer 9

yes.

Answer 10

\[\frac{2\log y + logy}{\log 2^2} = 6\]

Answer 11

where? @Mimi_x3

Answer 12

According to this, the answer is supposed to be 16. D:

Answer 13

\[\frac{3 \log y }{2\log 2} = 6\]
\[\frac32 \log(y-2) = 6\]

Answer 14

can u please guide what we will do after this? @Gab02282 .
the answer is 16.

Answer 15

I'm not sure, either. The question book just gave me the answer, but not the process. I'm sorry I can't be of any help. D:

Answer 16

no worries...
@Mimi_x3 im correct. just can't find steps to move forward.. :P
Do u know?

Answer 17

Make "6" an exponent? Since they have similar logs, can it be like, log (base 10) 10^6?

I don't know what I'm doing lol.

Answer 18

After that, eliminate the logs and solve for y. Right?

Answer 19

we have to solve for y.. yes..

Answer 20

//captain obvious
lol, I'm honestly unsure.

Answer 21

\[3\lg_{4}y/2\lg 2_{4} = 6\]
\[2\lg _{4}2 = 2*1/2\] = 1
=> \[3\lg _{4} y=6\]
=>\[\lg _{4}y = 2\]
\[y = 4^{2} = 16\]

Answer 22

@chemwile , mind showing the first two steps using draw tool?

Answer 23

Thank you very much, @chemwile!

Answer 24

vw