What am I doing wrong in this exponential equation?

[wahaj]

Homework Statement

5^2x - 3(5^x) = 10

Homework Equations

The Attempt at a Solution

log 5^2x - log 3(5^x) = log 10
2x log 5 - log 3 - x log 5 = log 10
x log 5 = log 10 + log 3
x = (log 10 + log 3)/log 5
putting this in my calculator I get 2.11
by inspection I can tell that the answer will be x = 1. So what am I doing wrong?

 

[dydxforsn]

I think if you do the substitution $u = 5^x$ you can solve a quadratic for 'u' and then get 'x' from that. One of the 2 solutions will be negative and therefore outside of the domain of the log.

Edit* Oh, well I guess I should say that the first line of your answer that you tried is already not right, you accidentally used the wrong properties for logs: http://dl.uncw.edu/digilib/mathematics/algebra/mat111hb/eandl/logprop/logprop.html

 

[SammyS]

Homework Statement

5^2x - 3(5^x) = 10

Homework Equations

The Attempt at a Solution

log 5^2x - log 3(5^x) = log 10
2x log 5 - log 3 - x log 5 = log 10
x log 5 = log 10 + log 3
x = (log 10 + log 3)/log 5
putting this in my calculator I get 2.11
by inspection I can tell that the answer will be x = 1. So what am I doing wrong?

First of all, log(a + b) ≠ log(a) + log(b)

To solve your equation, use the exponential property that, $\displaystyle \ \ 5^{2x}=\left(5^x\right)^2\ .$

Then letting u = 5^x, you will have a quadratic equation in u .

 

[wahaj]

oh that makes more sense. I thought I could do that. Well thanks for the help I got the answer I needed.

 

[Ray Vickson]

oh that makes more sense. I thought I could do that. Well thanks for the help I got the answer I needed.

No, you cannot say 'log of a sum = sum of the logs'---that is just false. For example, log(5+1) = log(6) (because 5+1=6), but log(5) + log(1) = log(5) (because log(1) = 0).