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!