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

  • Thread starter Thread starter cbarker1
  • Start date Start date
  • Tags Tags
    Logarithms
AI Thread Summary
To express the combination of logarithms as a single logarithm, start by applying the power rule to rewrite the expression as $\log_{d}(a^4) - \log_{d}(b^{5/6}) + \log_{d}(c^{2/3})$. Use the properties of logarithms: the sum of logarithms corresponds to multiplication, and the difference corresponds to division. The final expression can be simplified to $\log_{d}\left(a^4 \sqrt[6]{\frac{c^4}{b^5}}\right)$. The order of combining terms does not affect the final result, allowing flexibility in the approach.
cbarker1
Gold Member
MHB
Messages
345
Reaction score
23
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
 
Mathematics news on Phys.org
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.
 
Should I use the quotient to difference rule or the sum to multiplication rule first?The answer in the back of the book is $\log_{d}\left({a^4\sqrt[6]{\frac{c^4}{b^5}}}\right)$
 
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. :)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...

Similar threads

Replies
2
Views
2K
Replies
1
Views
4K
Replies
41
Views
5K
Replies
2
Views
10K
Replies
1
Views
10K
Replies
5
Views
2K
Replies
4
Views
11K
Back
Top