# Thread: How do you solve this logarithm?

1. ## How do you solve this logarithm?

If ln a = 2, ln b = 3, and ln c = 5, evaluate:

1. ln((a^-4)(b^1)) / ln(bc)^3

and

2. (ln c^-1)(ln a/(b^3))^4

2. Recall and apply your log rules.

log (AB) = log A + log B

log(A/B) = log A - log B

log (A)^n = n logA

3. In the following lines, I mean to say ln(a^-4), not ln(a-4). Latex issues, sorry.

$\displaystyle \frac{ln[(a^-4)(b)]}{ln(bc)^3}$

$\displaystyle \frac{ln(a^-4)+ln(b)}{3ln(bc)}$

$\displaystyle \frac{-4ln(a)+ln(b)}{3ln(b)+3ln(c)}$

$\displaystyle \frac{-4(2)+3}{3(3)+3(5)}$

$\displaystyle \frac{-5}{24}$

4. Originally Posted by rtblue
In the following lines, I mean to say ln(a^-4), not ln(a-4). Latex issues, sorry.

$\displaystyle \frac{ln[(a^-4)(b)]}{ln(bc)^3}$

$\displaystyle \frac{ln(a^-4)+ln(b)}{3ln(bc)}$

$\displaystyle \frac{-4ln(a)+ln(b)}{3ln(b)+3ln(c)}$

$\displaystyle \frac{-4(2)+3}{3(3)+3(5)}$

$\displaystyle \frac{-5}{24}$
FYI you can use $$\ln \left(a^{-4}\right)$$ which gives $\displaystyle \ln \left(a^{-4}\right)$

I've nothing new to contribute to the thread, look at the two posts above me