Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

So, I have this logarithmic expression $$\log_5 8-\log_5 20-\log_5 10$$ that I know how to evaluate - the quotient of numbers is the difference of logarithms, so you divide, etc. - but how the heck do you solve one like this with three logs?

I've been able to solve others that only had two logs, but this one confuses me. Help?

share|improve this question
log a - log b - log c = log (a/bc) –  Shahab Oct 7 '12 at 17:19
This is an expression which you want to evaluate, not an equation (no equal sign) to solve. –  Ross Millikan Oct 7 '12 at 17:49
Thanks for the grammatical contribution. It really helps. –  Brandt Oct 7 '12 at 18:03

4 Answers 4

up vote 1 down vote accepted

We know that $\log(a)- \log(b)= \log(\frac{a}{b})$. Hence, we have

$\log_5(8) - \log_5(20)-\log_5(10) = \log_5(\frac{8}{20})-\log_5(10) $

$= \log_5(\frac{\frac{8}{20}}{10})$

$= \log_5(\frac{8}{200})$

$= \log_5(\frac{1}{25})$

$= \log_5(1)-\log_5(25)$

$= 0 -2 = -2$

share|improve this answer
This is exactly how I ultimately solved it. Thanks. :) –  Brandt Oct 7 '12 at 23:18

A start: If you know how to handle the sum or difference of two logarithms, you can handle the sum, difference of arbitrarily many. Just deal with them two at a time. For example, $$\log_5 8-\log_5 20-\log_5 10=(\log_5 8 -\log_5 20)-\log_5 10=\log_5(8/20)-\log_5 10=\log_5(?).$$

Added: For a problem like this one, which involves smallish integers, there is another way to proceed. Note that $8=2^3$, so $\log_5 8=3\log_5 2$.

Similarly, $\log_5 20=\log_5(2^2\cdot 5)=2\log_5 2+\log_5 5=2\log_5 2+1$. Similarly, $\log_5 10=\log_5 2+1$. So our expression is equal to $$3\log_5 2-(2\log_5 2+1)-(\log_5 2+1).$$ Simplify. The $\log_5 2$ terms disappear.

share|improve this answer

$$\log_5 8-\log_5 20-\log_5 10=\log_5 8/200=\log_5 1/25=\log_5 1-\log_5 25=0-2=-2 $$

share|improve this answer

Implement the formula:

$1)$ $\log_x a-\log_x b=\log_x \frac{a}{b}$

$2)$ $\log_a 1=\log_a a^0=0$

$3)$ $\log_x a^n=n\log_x a$

$\log_5 8-\log_5 20-\log_5 10=\log_5\frac{8}{20}-\log_5 10=\log_5\frac{2}{5}-\log_5 10=\log_5\frac{\frac{2}{5}}{10}=\log_5 \frac{2}{50}=\log_5 \frac {1}{25}=\log_5 1 -\log_5 25=0-\log_5 5^2=0-2=-2$

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.