Logarithm Evaluating

Question

i'm new to this site and I need help on this logarithm question. I don't know how to approach this question to simplify it.

$\frac{\log _2\left(81\right)}{2-\log _2\left(18\right)}$

Apparently the answer is $-\frac{4\ln \left(3\right)}{\ln \left(\frac{9}{2}\right)}$

score 1 · Accepted Answer · answered Apr 25 '14 at 05:25

1

Use $\displaystyle\log_a(b^m)=m\log_ab$ when both logarithm remain defined unlike $\displaystyle\log_a1=\log_a(-1)^2=2\log_a(-1)$

and $\displaystyle\log_a(bc)=\log_ab+\log_ac$ when both logarithm remain defined unlike $\displaystyle\log_a6=\log_a(-3)(-2)=\log_a(-3)+\log_a(-2)$ for real $a>0$

Observe that $81=3^4$

answered Apr 25 '14 at 05:25

lab bhattacharjee

274,582

$\frac{4\log _2\left(3\right)}{2-\log _2\left(18\right)}$ What about the constant in the denominator? – user145591 Apr 25 '14 at 05:29
@user145591, $$\log_2{18}=\log_22+\log_2(3^2)=1+2\log_23$$ – lab bhattacharjee Apr 25 '14 at 05:30
$\frac{4\log _2\left(3\right)}{1-2\log _2\left(3\right)}$? – user145591 Apr 25 '14 at 05:35
@user145591, Have you understood the method? – lab bhattacharjee Apr 25 '14 at 05:37
Yes, is my answer there fully simplified? – user145591 Apr 25 '14 at 05:38

score 1 · Answer 2 · answered Apr 25 '14 at 06:39

You can start going directly from base $2$ to base $e$ and write $$\frac{\log _2\left(81\right)}{2-\log _2\left(18\right)}=\frac{\log (81)}{\log (2) \left(2-\frac{\log (18)}{\log (2)}\right)}=\frac{\log (81)}{2 \log (2)-\log (18)}=\frac{\log (81)}{\log (4)-\log (18)}$$ and continue simplifying since $$\log(81)=\log(3^4)=4 \log(3)$$ and $$\log (4)-\log (18)=-\log (\frac{18} {4})=-\log (\frac{9} {2})$$

Logarithm Evaluating

2 Answers2