# Solve for a variable in the exponent

1. Jan 21, 2009

### Bacat

I can't remember how to solve an equation of the following form:

$$ln(a^x) = ln(b^x)$$

I want to solve this for x where a and b are both real numbers. I know I can manipulate the logarithm to write:

$$xln(a) = xln(b)$$

But this just seems to cancel out the x's. If you can show me how to solve this type of equation for x, I would very much appreciate it (it's not homework, just trying to remember how to do it).

2. Jan 21, 2009

### NoMoreExams

As it should, ln(a^x) = ln(b^x) iff a = b

3. Jan 21, 2009

### Bacat

Good point. I think what I meant to say was:

$$C*ln(a^x) = D*ln(b^x)$$ where a,b,C,D are all real numbers.

This is currently what I'm thinking about it...

$$xC*ln(a) = xD*ln(b)$$
$$xC*ln(a) - xD*ln(b) = 0$$
$$x*(C*ln(a) - D*ln(b)) = 0$$

It seems close, but I can't quite find x.

Last edited: Jan 21, 2009
4. Jan 21, 2009

### Mentallic

Just factorize x out. If you ever feel like you want to divide through by the variable - stop - and rather factorize x out
This means x would be equal to 0 and possibly other values too. When you divide through by x, the denominator can never equal 0 therefore you lose your x=0 answer.

$$Cx ln a - Dx ln b=0$$

$$x(C ln a - D ln b)=0$$

5. Jan 21, 2009

### gabbagabbahey

I think you have one too many f's in that statement. It is also true for all 'a' and 'b' if x=0.

6. Jan 21, 2009

### Bacat

It still looks like x can be anything though. Maybe I'm still phrasing the problem wrong. I remember doing something like this a long time ago and x had a definite value...

$$ln(a^x)=ln(b^x)$$ where $$a=b$$