# How do you condense 20lna-4lnb?

Aug 23, 2016

$\ln \left({a}^{20} / {b}^{4}\right)$ or $4 \ln \left({a}^{5} / b\right)$.

#### Explanation:

$20 \ln a = \ln {a}^{20}$ by the rule $b \log a = \log {a}^{b}$.

Similarly, $4 \ln b = \ln {b}^{4}$ by the same rule.

So now we have $20 \ln a - 4 \ln b = \ln {a}^{20} - \ln {b}^{4}$.

Now, using the rule $\log a - \log b = \log \left(\frac{a}{b}\right)$, we can rewrite this as $\ln \left({a}^{20} / {b}^{4}\right)$.

If you like, you can write this as $\ln \left({\left({a}^{5} / b\right)}^{4}\right)$, which can be rewritten as $4 \ln \left({a}^{5} / b\right)$.

You can plug numbers in to these rules to make sure they work. For example, ${\log}_{2} 32 - {\log}_{2} 8 = 5 - 3 = 2$. If you use the division rule, you get ${\log}_{2} 32 - {\log}_{2} 8 = {\log}_{2} \left(\frac{32}{8}\right) = {\log}_{2} \left(4\right) = 2$.

More generally, we can show this rule and the others to be true. I will show that $\log a + \log b = \log \left(a \cdot b\right)$.

Start with $y = \ln a + \ln b$.

Exponentiate both sides to get ${e}^{y} = {e}^{\ln a + \ln b}$.

Through rules of exponents, the right side can be rewritten to get ${e}^{y} = {e}^{\ln} a \cdot {e}^{\ln} b$.

Since ${e}^{\ln} x = x$ (if you're confused why, say something) we get ${e}^{y} = a \cdot b$.

To finish, take the natural log of both sides to get $y = \ln \left(a \cdot b\right)$.

Similar things can be done to prove the other rules true.