Question

Let \[\log_{b}A=1 ; \log_{b}C=3 ; \log_{b}D=4 \] . What is the value of \[\log_{b}(\frac{ A^5C^2 }{ D^6 } )\] ?

A. 0
B. 26
C. -13
D. There isn't enough information to answer the question.

Answer 1

do you know the rule
\[ \log\left(\frac{a}{b}\right)= \log(a) - \log(b) \]
and
\[ \log(ab) = \log(a)+\log(b) \]
and
\[ \log(a^b) = b \log(a) \]

Answer 2

treat A^5 C^2 as one "thing" and use the first rule. what do you get ?

Answer 3

in other words, in the rule, replace little a with A^5 C^2 and b with D^6

Answer 4

this is my first time ever working with log so Im not sure if i did this write, but i got: a=(d6/5)/c2/5

Answer 5

@phi

Answer 6

the idea to using the rule, is "pattern match". in other words, if you have the rule
\[ \log\left(\frac{a}{b}\right)= \log(a) - \log(b)\]
and we want to use it for
\[ \log_{b}\left(\frac{ A^5C^2 }{ D^6 } \right) \]
we match "a" with the stuff on top: A^5C^2
and "b" in the rule with the stuff on the bottom
using the rule we get
\[ \log_{b}\left(\frac{ A^5C^2 }{ D^6 } \right) = \log_{b}(A^5C^2) - \log_b(D^6)\]

Answer 7

If that is not clear, we have another chance, because now we can use the 2nd rule for the first term.
use
\[ \log(ab) = \log(a)+\log(b) \]
on
\[ \log_{b}(A^5C^2) \]
match a with A^5 and b with C^2
and rewrite the log expression as log(a) + log(b) (but with A^5 replacing "a" and C^2 replace "b")