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Here is a equation:

$$ a^b = ab$$

[ that is, $a$ raised to $b$ is equal to $a$ times $b$ ]

Find all real values of $a$ and $b$ where $a$ is not equal to zero, $b$ is not equal to one.

[Difficulty] Two variables can be solved only if two distinct equations are provided describing the relationship between them. However, here only one equation is provided in which two variables exist and the challenge is to find all such sets of real numbers that satisfy the given relationship.

Hint: Infinite many such values satisfy the above

[Thoughts] Let $b = a^k$

then the given equation becomes

$$a^{a^k} = a \cdot a^k$$

or $a^k = k+1$ which means, $a^k - k = 1$ for all $a$ and $k$ where $a$ is a variable and $k$ can be a constant starting from say $0$.

Let $a =1$, $k = 0$ which satisfies the above equation $1^0 - 0 =1$. Accordingly $a=1$, $b=1$ Let $a =2$, $k = 1$ then $2 ^1 - 1 = 1$ also satisfy the condition. Accordingly $a=2$, $b=1$ Let $a=\sqrt3$ , $k=2$, then $(\sqrt3)^2 - 2 = 1$ and accordingly $a= \sqrt{3}$, $b = 3$

Another set of such numbers that satisfy the above condition is $\frac{9}{4}$ and $\frac{3}{2}$ as $(\frac{9}{4})^{\frac{3}{2}} = \frac{9}{4}\cdot\frac{3}{2}$.

In general, all $a$ and $k$ where $a= p^{\frac{1}{n}}$ and $k=n$ satisfy the above condition for all $p \in \mathrm{R}$.

Please check this and let me know if my approach is correct. Thanks.

In general, all $a$ and $k$ where $a=p^{\frac{1}{n}}$ and $k=n$ satisfy the above condition for all $p \in \mathrm{R}$.

I mean,

For any real number $p$, there is an $n$ such that $a = p^{\frac{1}{n}}$ and $b = a^n$ will be a solution.

Please check this and let me know if my approach is correct. Thanks.

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  • $\begingroup$ You decided to try expressing $b$ as a power $a^k$ of $a$. You then get somewhat distracted with arguing special cases, and finally you introduce a new variable $p$. But if we are dealing with $p \gt 0$, then $a = p^{\frac{1}{n}}$ and $b = a^n$. Thus $p=b$ and we get nothing new. Try letting $a$ be a power of $b$, and the Answer of TonyK should appear quickly. $\endgroup$ – hardmath Sep 20 '14 at 14:05
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Given any positive $b \ne 1$, just let $a = b^{1/(b-1)}$.

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