How Can I Express a Combination of Logarithms as a Single Logarithm?

[cbarker1]

Express as a Single Logarithm. Assume all the logarithms have the same base. The problem is:
$4*\log_{d}\left({a}\right)-\frac{5}{6}*\log_{d}\left({b}\right)+\frac{2}{3}*\log_{d}\left({c}\right)$, where $d\gt1$

then I use the power rule for logarithm for the beginning and use some grouping symbols for last two:

$\log_{d}\left({a^4}\right)-(\frac{5}{6}\log_{d}\left({b}\right)+\frac{2}{3}\log_{d}\left({c}\right))$

Afterwards, I combine the coeffecients of the logarithms to be the common denominator of one sixth :
$\log_{d}\left({a^4}\right)-(\frac{5}{6}\log_{d}\left({b}\right)+\frac{4}{6}\log_{d}\left({c}\right))$

Here is where I have some trouble. Do I factor the one-sixth;then, use the sum-multiplication inside the group symbol or use the power again and the sum-multiplication rule for log inside the grouping symbol like this:
$\log_{d}\left({a^4}\right)-(\frac{5}{6}\log_{d}\left({b}\right)+\frac{4}{6}\log_{d}\left({c}\right))$ to
$\log_{d}\left({a^4}\right)-(\frac{1}{6}({5}\log_{d}\left({b}\right)+4\log_{d}\left({c}\right)))$
from that to this
$\log_{d}\left({a^4}\right)-(\frac{1}{6}(\log_{d}\left({b^5*c^4}\right))$
to $\log_{d}\left({a^4}\right)-\frac{1}{6}(\log_{d}\left({b^5*c^4}\right))$

Or

$\log_{d}\left({a^4}\right)-(\frac{5}{6}\log_{d}\left({b}\right)+\frac{4}{6}\log_{d}\left({c}\right))$
$\log_{d}\left({a^4}\right)-(\log_{d}\left({\sqrt[6]{b^5}}\right)+\log_{d}\left({\sqrt[6]{c^4}}\right))$
$\log_{d}\left({a^4}\right)-(\log_{d}\left({\sqrt[6]{b^5}*\sqrt[6]{c^4}}\right))$
$\log_{d}\left({a^4}\right)-(\log_{d}\left({\sqrt[6]{b^5*c^4}}\right))$

What to do next?Thank you

Cbarker1

 

[Jameson]

I think you can skip a lot of the intermediate steps you are taking. It's not wrong the way you're doing it though.

$$4*\log_{d}\left({a}\right)-\frac{5}{6}*\log_{d}\left({b}\right)+\frac{2}{3}*\log_{d}\left({c}\right)$$

$$\log_{d}\left({a^4}\right)-\log_{d}\left({b}^{\frac{5}{6}} \right)+\log_{d}\left({c}^{\frac{2}{3}}\right)$$

From here you need to remember two rules:

1) $$\log_{d}\left({a}\right)+\log_{d}\left({b}\right)=\log_{d}\left({ab}\right)$$

2) $$\log_{d}\left({a}\right)-\log_{d}\left({b}\right)=\log_{d}\left({\frac{a}{b}}\right)$$

Combine any two terms into one, then combine that with the third term. Simplify if possible at the end and keep powers written in fractional form until the end too.

 

[cbarker1]

Should I use the quotient to difference rule or the sum to multiplication rule first?The answer in the back of the book is $\log_{d}\left({a^4\sqrt[6]{\frac{c^4}{b^5}}}\right)$

 

[Jameson]

It doesn't matter. You'll get the same answer either way, which is why I wrote pick any two terms to combine. If it's easier for you to stick with terms left to right then try combining the first two and see what you get. :)