SOLUTION: Im having difficulty with solving the following logarithim for X 6^(2log base6 X + log base6 X) = 125 I thought the log base of 6 would cancel out and I would get x^2 +

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Im having difficulty with solving the following logarithim for X 6^(2log base6 X + log base6 X) = 125 I thought the log base of 6 would cancel out and I would get x^2 +      Log On


   

Question 633487: Im having difficulty with solving the following logarithim for X
6^(2log base6 X + log base6 X) = 125

I thought the log base of 6 would cancel out and I would get
x^2 + X =5^3
but what do I do with that 6? add it to get X^2 + X +6 =125?
please help thank you
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Im having difficulty with solving the following logarithim for X
6^(2log base6 X + log base6 X) = 125
**
6^(2log base6 X + log base6 X) = 125
place under single log
6^(log6[(x^2)x])=125
6^(log6[x^3])=125
base(6) raised to log of number(x^3)=number(x^3) (definition of logarithm)
x^3=125=5^3
x=5
..
Check: using calculator and change of base formula:
2log base6 X + log base6 X=2log base6(5) + log base6(5)≈2.694733
6^2.694733≈124.999954..