Solve for a variable in the exponent

In summary, the conversation is about solving an equation of the form ln(a^x) = ln(b^x) for x, where a and b are real numbers. The individual is aware of manipulating the equation to xln(a) = xln(b), but is unsure of how to proceed from there. They suggest possibly factoring out x, but are unsure if that is the correct approach. The expert advises against dividing through by x and instead suggests factoring it out to find the solution of x=0. The individual realizes that x can be any value if a=b, but the only true value for x for any real value of a and b is 0.
  • #1
Bacat
151
1
I can't remember how to solve an equation of the following form:

[tex]ln(a^x) = ln(b^x)[/tex]

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

[tex]xln(a) = xln(b)[/tex]

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).

Thank you for your time.
 
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  • #2
As it should, ln(a^x) = ln(b^x) iff a = b
 
  • #3
Good point. I think what I meant to say was:

[tex]C*ln(a^x) = D*ln(b^x)[/tex] where a,b,C,D are all real numbers.

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

[tex]xC*ln(a) = xD*ln(b)[/tex]
[tex]xC*ln(a) - xD*ln(b) = 0[/tex]
[tex]x*(C*ln(a) - D*ln(b)) = 0[/tex]

It seems close, but I can't quite find x.
 
Last edited:
  • #4
Just factorize x out. If you ever feel like you want to divide through by the variable - stop - and rather factorize x out :wink:
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.

[tex]Cx ln a - Dx ln b=0[/tex]

[tex]x(C ln a - D ln b)=0[/tex]
 
  • #5
NoMoreExams said:
As it should, ln(a^x) = ln(b^x) iff a = b

I think you have one too many f's in that statement.:wink: It is also true for all 'a' and 'b' if x=0.
 
  • #6
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...

Thanks for your help anyway.
 
  • #7
Bacat said:
It still looks like x can be anything though. Maybe I'm still phrasing the problem wrong.
It would look like (and be) true for all x if:

[tex]ln(a^x)=ln(b^x)[/tex] where [tex]a=b[/tex]

But the only true value for x for any real value of a and b is 0.
 

Related to Solve for a variable in the exponent

1. How do I solve for a variable in the exponent?

To solve for a variable in the exponent, you can use logarithms, specifically the logarithm with the base equal to the base of the exponent. Take the logarithm of both sides of the equation and then use the laws of logarithms to isolate the variable.

2. Can I solve for a variable in the exponent without using logarithms?

Yes, you can also use the properties of exponents to solve for a variable in the exponent. For example, if the variable is in the exponent on both sides of the equation, you can set the exponents equal to each other and solve for the variable.

3. What if there are multiple variables in the exponent?

If there are multiple variables in the exponent, you can solve for one variable at a time by treating the other variables as constants. You may need to rearrange the equation and use logarithms to isolate each variable.

4. Is it possible to have multiple solutions when solving for a variable in the exponent?

Yes, it is possible to have multiple solutions when solving for a variable in the exponent. This can happen when the base of the exponent is a fraction or when there are multiple terms with the variable in the exponent on one side of the equation.

5. Are there any special cases when solving for a variable in the exponent?

There are a few special cases to keep in mind when solving for a variable in the exponent. These include when the exponent is a negative or a fractional power, and when the equation contains imaginary numbers. In these cases, you may need to use more advanced techniques to find the solution.

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