Question 57683
The easiest way to solve this is to change the bases before you take the logs:
{{{25^(x+5)=125^(x+4)}}}  25=5^2, and 125=5^3
{{{5^(2(x+5))=5^(3(x+4))}}}  Now that their bases are the same you can equate their exponents:
2(x+5)=3(x+4)
2x+10=3x+12
2x-2x+10=3x-2x+12
10=x+12
10-12=x+12-12
-2=x
Check:
{{{25^(-2+5)=125^(-2+4)}}}
{{{25^3=125^2}}}
{{{15625=15625}}}  We're right.
:
Since you asked to take the logs of both sides, I will show you that method, but it ain't gonna be pretty.
{{{25^(x+5)=125^(x+4)}}}
{{{log(25^(x+5))=log(125^(x+4))}}}
{{{(x+5)log(25)=(x+4)log(125)}}}
{{{xlog(25)+5log(25)=xlog(125)+4log(125)}}}
{{{xlog(25)-xlog(125)+5log(25)-5log(25)=xlog(125)-xlog(125)+4log(125)-5log(25)}}}
{{{x(log(25)-log(125))=log(125^4)-log(25^5)}}}
{{{xlog(25/125)=log(125^4/25^5)}}}
{{{xlog(1/5)=log(25)}}}
{{{xlog(1/5)/log(1/5)=log(25)/log(1/5)}}}  You'll need a scientific calculator here.  The other way didn't require one.
{{{x=-2}}}
:
Whatever makes your teacher happy.
Happy Calculating!!!