Question

express as a single log:
1/2log[5]x+2log[5](x+2)-[log[5](1+x)+log[5](1-x)]

Answer 1

there are three rules you need here:\[\log(u)+\log(v)=\log(uv)\]\[\log(u)-\log(v)=\log \left( \frac{u}{v} \right)\]and\[r*\log(u)=\log \left( u^r \right)\]

Answer 2

using the third rule twice
\[\log_{5}(x^{1/2})+\log-{5}(x+2)^2−[\log_{5}(1+x)+\log_{5}(1+x)] \]apply rule number 1 twice
\[=\log_{5}(x^{1/2}(x+2)^2)−[\log_{5}(1−x)(1+x)] \]applying rule 2 and simplifying the argument of the second log
\[=\log_{5} \frac{x^{1/2}(x+2)^2}{(1−x^2)}\]

Answer 3

i see an obvious typo in the first algebraic line of this post the log_{5} looks like a minus 5; just rewrite that -5 as a base for the log