How do you condense 4[ln z+ln(z+5)-2ln(z-5)]?

Jul 2, 2018

$4 \ln \left(\frac{z \left(z + 5\right)}{z - 5} ^ 2\right)$

Explanation:

We have the following:

$4 \left(\textcolor{b l u e}{\ln z + \ln \left(z + 5\right)} - 2 \ln \left(z - 5\right)\right)$

Recall the logarithm/natural log rule

$\ln a + \ln b = \ln \left(a b\right)$

This allows us to rewrite the blue terms as

$4 \left(\textcolor{b l u e}{\ln \left(z \left(z + 5\right)\right)} - \textcolor{p u r p \le}{2 \ln \left(z - 5\right)}\right)$

We can reference another logarithm/natural log rule:

$a {\log}_{c} b = {\log}_{c} \left({b}^{a}\right)$

We can apply this to the purple term. Our coefficient simply becomes our exponent. We now have

$4 \left(\textcolor{b l u e}{\ln \left(z \left(z + 5\right)\right)} - \textcolor{p u r p \le}{\ln {\left(z - 5\right)}^{2}}\right)$

We can leverage yet another logarithm/natural log rule:

${\log}_{c} a - {\log}_{c} b = {\log}_{c} \left(\frac{a}{b}\right)$

If we have the same base and we're subtracting, we can turn this into division. We now have

$4 \ln \left(\frac{\textcolor{b l u e}{z \left(z + 5\right)}}{\textcolor{p u r p \le}{{\left(z - 5\right)}^{2}}}\right)$

Hope this helps!