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

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Answer 1 · 2018-07-02T23:38:20

$4ln((z(z+5))/(z-5)^2)$

We have the following:

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

Recall the logarithm/natural log rule

$lna+lnb=ln(ab)$

This allows us to rewrite the blue terms as

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

We can reference another logarithm/natural log rule:

$alog_cb=log_c(b^a)$

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

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

We can leverage yet another logarithm/natural log rule:

$log_ca-log_cb=log_c(a/b)$

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

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

Hope this helps!

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