Question

Write the expression as a single logarithm.

3 logvb q + 6logvb v

Answer 1

\[\Large\bf\sf 3\log vb q+6\log vbv\]

Is that what it looks like?
Or is one of those letters supposed to be an exponent or base of the log maybe?

Answer 2

B is a base i think for both 3 and 6

Answer 3

\[3\log_{b}q + 6\log_{b}V \]

Answer 4

Here is a rule of logs that will allow us to deal with coefficients in front of the log.
\[\Large\bf\sf \color{orangered}{b}\cdot \log(a)\quad=\quad \log(a^{\color{orangered}{b}})\]
Do you see how we can apply this rule to the 3 and 6?

Answer 5

Yeah i think so I haven't really used the log stuff

Answer 6

For the first term it will give us:\[\Large\bf\sf 3 \log_b q\quad=\quad \log_b q^3\]How bout for the other one?

Answer 7

it would be \[\log_{b}q ^{6} \] right?

Answer 8

Hmm the inside of our second log is V, not q, right? :o

Answer 9

oh yeah sorry xD

Answer 10

:3

Answer 11

\[\Large\bf\sf \log_b q^3+\log_b V^6\]
Ok good that brings us here.

Answer 12

Our other rule of logs that we'll need to apply:\[\Large\bf\sf \log(a)+\log(b)\quad=\quad \log(a\cdot b)\]The sum of logs can be written as a single log with the product of their arguments inside.

Answer 13

\[\Large\bf\sf \log_b (q^3)+\log_b (V^6)\]So how can we apply that here? :o

Answer 14

\[\log (q*v) or \log (3*6)\] ???

Answer 15

Hmm I'm not sure why you're taking things apart like that.
\[\Large\bf\sf a=q^3, \quad b=V^6\]

\[\Large\bf\sf \log_b (q^3)+\log_b (V^6)\quad=\quad \log_b(q^3\cdot V^6)\]

Answer 16

oh ok that makes more sense.. like i said i haven't really used log stuff before

Answer 17

ya fair enough D:

Answer 18

XD anyway lets continue im writing this stuff down so I can remember where to look :3

Answer 19

That would be your final answer.
We can't get it any more compact than that :o\[\Large\bf\sf \log_b q^3V^6\]

Answer 20

Awesome! Sorry i had to bring up groceries x3 Thank you for the help :D