Do equal bases imply equal powers?

Question

$$x^a = x^b \Rightarrow a =b$$

So, this is a concept I used in multiple math problems and they often turn out right. The thing is, today my math teacher told me that this is not necessarily true.
_{(He did not, however, give me a proper explanation as to why that is and no one expected him to because it seemed very trivial and seemed like something that everybody should have already known.)}

I was wondering if someone could explain as to why he said that. I presume it has something to do with higher levels of math that I don't understand.

My logic is that since $\log_{x} a = \log_{x} b$, $a = b$.
But that's only true if $f(a) = f(b) \Rightarrow a = b$.
I only assume so because I did that in many trigonometry questions. But I don't believe that's enough proof to substantiate my claim. Please help.

Answer 1

It works, provided that $x>0$ and $x\neq 1$ (these are the allowed values for a logarithm of base $x$). Otherwise, here are some counterexamples:

• $(-1)^3 = -1 = (-1)^5 \qquad\text{ yet }\qquad 3 \neq 5$
• $0^3 = 0 = 0^5 \qquad\text{ yet }\qquad 3 \neq 5$
• $1^3 = 1 = 1^5 \qquad\text{ yet }\qquad 3 \neq 5$

Note that if $x<0$ and $x \neq -1$, we can actually get rid of the negative sign so that it works. For example, if $(-2)^a = (-2)^b$, then taking the absolute value of both sides yields $2^a = 2^b$, so we may take the log (base $2$) of both sides to obtain $a=b$.

Answer 2

If you take logs in your original equation, then you obtain:

$x^a = x^b \\ \Rightarrow a\log x = b\log x \\ \Rightarrow (a-b)\log x = 0$

From this last equation, we see that either $a-b=0$, meaning $a=b$, or else $\log x = 0$, meaning $x=1$.

Other exceptions can occur for values of $x$ that aren't in the domain of the logarithm function at all. Dealing with negative values of $x$ can be done by canceling out any negatives before taking logs.

If you're allowing for complex numbers, the answer is somewhat more complicated. For example, $e^{2\pi i} = e^0$, but $2\pi i \neq 0$.

Answer 3

Whether the stated rule is valid depends on what $x$ is, and on where the values $a,b$ are supposed to live. The most important cases where the rule holds are when $x$ in an indeterminate (in which case it does not much matter where $a,b$ are taken): if the monomials $x^a$ and $x^b$ are the same, then $a$ and $b$ must be the same. In the same vein, if $x^a$ and $x^b$ designate functions of say $x\in\Bbb R_{>0}$ (and $a,b$ are constants, expressions not containing$~x$), then the equality of functions $x^a=x^b$ (more properly $(x\mapsto x^a)=(x\mapsto x^b)$) implies $a=b$ (you can prove this by differentiating both sides and then putting $x=1$ in the derivatives).

Howerever if $x$ designates some concrete (though possibly unspecified) value, then the implication is in general not valid, as has been illustrated by many examples (the case $x=1$ is the most spectacular example of failure). Under some specific conditions the rule will still be valid (such as $x\in\Bbb R_{>1}$ and $a,b\in\Bbb R$), but you need to be very explicit about these conditions.

A lesson to learn from this is that an equation like $x^a=x^b$ does not have an unambiguous meaning in isolation, and it needs to be accompanied by an indication of what the symbols mean.

Answer 4

That should be true if x does not equal to 0, 1 or -1.

Answer 5

Given $x^a = x^b$;

Then $\frac{x^a}{x^b} = x^{(a-b)} = 1$ ;

Let $a-b = n$ ;

Then $x = 1^{(1/n)} = 1^{1/n}$ ;

So $a=b$ only if $x$ is a real number , not one , not zero , not infinite.