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.

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

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

From here you need to remember two rules:

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

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

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. :)

 

[CellCoree]

,

I would first like to commend you on your approach to solving this problem. You correctly used the power rule for logarithms to rewrite the expression and then combined the coefficients to have a common denominator of one sixth.

To simplify the expression further, you can use the product rule for logarithms, which states that $\log_a(x*y) = \log_a(x) + \log_a(y)$. This means that you can rewrite the expression inside the parentheses as $\log_d\left(\sqrt[6]{b^5*c^4}\right)$, which is equivalent to $\log_d\left(\sqrt[6]{b^5}\right) + \log_d\left(\sqrt[6]{c^4}\right)$.

Then, using the power rule again, you can rewrite this as $\log_d(b^\frac{5}{6}) + \log_d(c^\frac{4}{6})$. Finally, combining the coefficients once more, you will get $\log_d(b^\frac{5}{6}*c^\frac{4}{6})$, which is equivalent to $\log_d\left(\sqrt[6]{b^5*c^4}\right)$.

Therefore, your final expression will be $\log_d(a^4) - \frac{1}{6}*\log_d\left(\sqrt[6]{b^5*c^4}\right)$.

I hope this helps and keep up the good work!