# Real analysis of powers

Show that if $a,b$ are rational numbers and $x$ is a positive real number then $x^a$$x^b = x^{(a+b)} I honestly have no idea how to even do this. Anyone have any hints or a good explanation? Thanks in advance! • note that x^a=e^{a\ln(x)} – Kamster Nov 7 '14 at 2:46 • What is your definition of exponentiation? – Jose Antonio Nov 7 '14 at 2:47 • Since he restricts this to rational numbers, perhaps log is not yet known? – GEdgar Nov 7 '14 at 2:47 • yea youre probably right GEdgar – Kamster Nov 7 '14 at 2:48 • Maybe first prove the case where a and b are integers. Then apply that with x replaced by x^{1/n}. – GEdgar Nov 7 '14 at 2:48 ## 3 Answers We know that$$x^ax^b=x^{a+b}$$is true for positive integer powers a,b and also it can be easily extend to negative integers also using$$x^{-a}=\dfrac{1}{x^a}$$where x is a positive real number. Note that using usual rules for integer exponents we can prove that,$$\Big(x^{\frac{p_1}{q_1}}x^{\frac{p_2}{q_2}}\Big)^{q_1q_2}=\Big(x^{\frac{p_1}{q_1}}\Big)^{q_1q_2}\Big(x^{\frac{p_2}{q_2}}\Big)^{q_1q_2}=x^{p_1q_2}x^{p_2q_1}=x^{p_1q_2+p_2q_1}.$$Hence taking the \dfrac{1}{q_1q_2} th power of both sides,$$x^{\frac{p_1}{q_1}}x^{\frac{p_2}{q_2}}=x^{\frac{p_1q_2+p_2q_1}{q_1q_2}}=x^{\frac{p_1}{q_1}+\frac{p_2}{q_2}}.$$The proof depends on the definition of x^a you are using. The easiest approach to defining x^a is to define, \log{y}=\int_1^y\frac{dt}{t} and use the inverse function theorem to define e^x to be the inverse function. Finally x^a=e^{a\log{x}}. You can now derive the rules of exponents from the rules of logarithms. The rules for logarithms are proved from the properties of the integral. The second approach is that you know exactly what x^n means when n is an integer. Assuming x is positive you use the inverse function theorem to define x^{1/n}. For any rational number x^{m/n}=(x^m)^{1/n}. Next you define, at least for a positive,$$x^a=\sup{\{x^{m/n}| 0<m/n<a\}}$$. You now prove the rules of exponents for integers by induction, and for roots by the inverse function theorem and the rules for integers. Finally you get the rules of exponents for real numbers via continuity of$x^a\$ and the rules of exponents for rational numbers.

I prefer the first path.

• the second path is similar to what we are doing in class – Zoë Soriano Nov 7 '14 at 3:13

The other answer took the calculus based approach, so I'll show you the algebraic approach to this proof.

First, take the log base x of both sides:

log[(x^a)(x^b)]/log(x) = log[x^(a+b)]/log(x)


Then simplify using log properties:

[log(x^a)+log(x^b)]/log(x) = log[x^(a+b)]/log(x)
[a*log(x)+b*log(x)]/log(x) = (a+b)log(x)/log(x)


Factoring out log(x) from the left side:

(a+b)log(x)/log(x) = (a+b)log(x)/log(x)


And thus:

a+b = a+b


The original statement must be true for x in the domain of the log function, so the only constraints are that x => 0, and that a and b are rational real numbers, exactly as your original statement dictated.