Question 254610
Here is the original problem:
{{{(1/2)*(3log(w-4)+2log(w+5)-log(7))}}}
First take the 3 and 2 up as exponents [power rule] as
{{{(1/2)*(log(w-4))^3 + (log(w+5))^2-log(7)}}}
Second, multiply log(w-4))^3 + (log(w+5))^2 to get
{{{(1/2)*(log(w-4))^3*((w+5))^2-log(7)}}}
Third, divide by log(7) to get
{{{((1/2)*(log(w-4))^3*((w+5))^2)/(7)}}}
Last bring the front 1/2 as a power as
{{{((log(w-4))^3*((w+5))^2/(7))^(1/2)}}}