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

Question

3998 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-02T23:38:20

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

We have the following:

$4 (ln z + ln (z + 5) - 2 ln (z - 5))$ $4(color(blue)(lnz+ln(z+5))-2ln(z-5))$

Recall the logarithm/natural log rule

$ln a + ln b = ln (a b)$ $lna+lnb=ln(ab)$

This allows us to rewrite the blue terms as

$4 (ln (z (z + 5)) - 2 ln (z - 5))$ $4(color(blue)(ln(z(z+5)))-color(purple)(2ln(z-5)))$

We can reference another logarithm/natural log rule:

$a {log}_{c} b = {log}_{c} (b^{a})$ $alog_cb=log_c(b^a)$

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

$4 (ln (z (z + 5)) - {ln (z - 5)}^{2})$ $4(color(blue)(ln(z(z+5)))-color(purple)(ln(z-5)^2))$

We can leverage yet another logarithm/natural log rule:

${log}_{c} a - {log}_{c} b = {log}_{c} (\frac{a}{b})$ $log_ca-log_cb=log_c(a/b)$

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

$4 ln (\frac{z (z + 5)}{{(z - 5)}^{2}})$ $4ln((color(blue)(z(z+5)))/(color(purple)((z-5)^2)))$

Hope this helps!

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