SOLUTION: How do you solve 25^(x+5)=125^(x+4) I think you are supposed to use the change of base formula? I keep getting confused when I take the log of both sides. Hope you can help. Thanks

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: How do you solve 25^(x+5)=125^(x+4) I think you are supposed to use the change of base formula? I keep getting confused when I take the log of both sides. Hope you can help. Thanks      Log On


   

Question 57683: How do you solve 25^(x+5)=125^(x+4) I think you are supposed to use the change of base formula? I keep getting confused when I take the log of both sides. Hope you can help. Thanks
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
The easiest way to solve this is to change the bases before you take the logs:
25%5E%28x%2B5%29=125%5E%28x%2B4%29 25=5^2, and 125=5^3
5%5E%282%28x%2B5%29%29=5%5E%283%28x%2B4%29%29 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%5E%28-2%2B5%29=125%5E%28-2%2B4%29
25%5E3=125%5E2
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%5E%28x%2B5%29=125%5E%28x%2B4%29
log%2825%5E%28x%2B5%29%29=log%28125%5E%28x%2B4%29%29
%28x%2B5%29log%2825%29=%28x%2B4%29log%28125%29
xlog%2825%29%2B5log%2825%29=xlog%28125%29%2B4log%28125%29

x%28log%2825%29-log%28125%29%29=log%28125%5E4%29-log%2825%5E5%29
xlog%2825%2F125%29=log%28125%5E4%2F25%5E5%29
xlog%281%2F5%29=log%2825%29
xlog%281%2F5%29%2Flog%281%2F5%29=log%2825%29%2Flog%281%2F5%29 You'll need a scientific calculator here. The other way didn't require one.
x=-2
:
Whatever makes your teacher happy.
Happy Calculating!!!