0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Do you know the basic log rules? Laid out here: purplemath.com/modules/logrules.htm $\endgroup$ Commented Mar 18, 2015 at 20:22
  • $\begingroup$ not really have lots of questions to do like this. if i am shown how to answer this then i have a referance $\endgroup$ Commented Mar 18, 2015 at 20:25
  • $\begingroup$ No i wrote it exactly how i got it, just unable to format it. $\endgroup$ Commented Mar 18, 2015 at 20:28
  • $\begingroup$ You will get very far just using the 3 rules on the link above. There are answers below that show how to use them. $\endgroup$ Commented Mar 18, 2015 at 20:29

2 Answers 2

2
$\begingroup$

$2\log_b6+\frac12\log_b25-\log_b30=$

$\log_b6^2+\log_b25^\frac12-\log_b30=$

$\log_b36+\log_b5-\log_b30=$

$\log_b\frac{36\cdot5}{30}=$

$\log_b6$

$\endgroup$
1
$\begingroup$

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 $25^\frac{1}{2}=\sqrt{25}=5$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .