How do you simplify a logarithm expression with multiple terms?

[Shawn Garsed]

Homework Statement

Simplify x^(3logx2 - logx5) to find an exact numerical value.

Homework Equations

The Attempt at a Solution

3logx2=logx2^3 or logx8,
(logx8 - logx5)=logx8/5
the inverse would be x^y=8/5 (y is unknown)
therefore logx8/5=y and x^(logx8/5)=x^y=8/5
and the inverse of that would be logx8/5=logx8/5

(logx8/5)-(logx8/5)=0
rewriting it you get logx(8/5)/(8/5) or just logx1=0
once again the inverse would be 1=x^0 which tells me that x can be any given number.
Is this correct?

 

[LCKurtz]

Homework Statement

Simplify x^(3logx2 - logx5) to find an exact numerical value.

Homework Equations

The Attempt at a Solution

3logx2=logx2^3 or logx8,
(logx8 - logx5)=logx8/5
the inverse would be x^y=8/5 (y is unknown)
therefore logx8/5=y and x^(logx8/5)=x^y=8/5
and the inverse of that would be logx8/5=logx8/5

(logx8/5)-(logx8/5)=0
rewriting it you get logx(8/5)/(8/5) or just logx1=0
once again the inverse would be 1=x^0 which tells me that x can be any given number.
Is this correct?

Your question asked you to just simplify the expression. You have shown that the expression given reduces to 8/5, which is correct, and you could stop there. And you are also correct that it doesn't matter what x is, although you would want x > 0.

 

[Shawn Garsed]

Sorry for the late response, for some reason I couldn't login to my account yesterday.

Your question asked you to just simplify the expression. You have shown that the expression given reduces to 8/5, which is correct, and you could stop there.

Your right, I was trying to find an exact numerical value for x.

And you are also correct that it doesn't matter what x is, although you would want x > 0.

Shouldn't it be x>1, since 1^y always equals 1.

 

[Mark44]

Right, the base shouldn't be 1 either. So the conditions are x > 0 and x $\neq$ 1.