# rewrite logarithmic expression

I have this logarithmic expression

2 logb 6 + (1/2) logb 25 - logb 30

and have to rewrite it as logb of one number. I just don't understand how to do this. help please.

• Do you know the basic log rules? Laid out here: purplemath.com/modules/logrules.htm Commented Mar 18, 2015 at 20:22
• not really have lots of questions to do like this. if i am shown how to answer this then i have a referance Commented Mar 18, 2015 at 20:25
• No i wrote it exactly how i got it, just unable to format it. Commented Mar 18, 2015 at 20:28
• You will get very far just using the 3 rules on the link above. There are answers below that show how to use them. Commented Mar 18, 2015 at 20:29

$2\log_b6+\frac12\log_b25-\log_b30=$
$\log_b6^2+\log_b25^\frac12-\log_b30=$
$\log_b36+\log_b5-\log_b30=$
$\log_b\frac{36\cdot5}{30}=$
$\log_b6$
I assume you mean: $$2 \log_b(6) + \frac{1}{2}\log_b(25) - \log_b(30).$$ Then just apply the basic rules i.e. $\log_b(a)+\log_b(c)=\log_b(a\cdot c)$, $\quad c\log_b(a) = \log_b(a^c)$ and $\log_b(a)-\log_b(c) = \log_b\left( \frac{a}{c} \right)$. Now easily we see that it must equal $$\log_b(6^2) + \log_b(25^{\frac{1}{2}}) - \log_b(30) = \log_b\left(\frac{36\cdot 5}{30}\right)=\log_b(6)$$ Which is $\log_b(6)$ as $6^2=36$ and $25^\frac{1}{2}=\sqrt{25}=5$.