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According to wikipedia, we have that

$$ x^{\log_b(y)} = y^{\log_b(x)} $$

because

$$ x^{\log_b(y)} = b^{\log_b(x) \log_b(y)} = b^{\log_b(y) \log_b(x)} = y^{\log_b(x)} $$

But what justifies that first leap?

$$ x^{\log_b(y)} = b^{\log_b(x) \log_b(y)} $$

PP1211
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5 Answers5

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It starts from the fact that $x$ is equal to $b^{\log_b(x)}$. To elaborate, the log to the base $b$ of $x$ is defined as the number which, when $b$ is raised to this number, gives back $x$. Now, since $x = b^{\log_b(x)}$, we can always replace any occurrence of $x$ in any equation whatsoever with $b^{\log_b(x)}$. So, breaking that first leap into two steps, we first have:

$x^{log_b(y)} = (b^{\log_b(x)})^{log_b(y)}$

From here, we can apply a known rule of exponents which is that $(s^r)^t = s^{rt}$. Applying that rule we get:

$(b^{\log_b(x)})^{log_b(y)} = b^{\log_b(x)log_b(y)}$

Which is the first step in the above.

Colm Bhandal
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Notice that $\def\lfrac#1#2{{\large\frac{#1}{#2}}}$$a^b=e^{b\ln(a)}$ and $\log_c (d)=\lfrac{\ln(d)}{\ln(c)}$. Therefore:

$$x^{\log_b(y)}=e^{\log_b(y)\ln(x)}=e^{\lfrac{\ln(y)}{\ln(b)}\ln(x)}=e^{\ln(y)\lfrac{\ln(x)}{\ln(b)}}=e^{\ln(y)\log_b(x)}=y^{\log_b(x)}$$

user21820
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Surb
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    Your original had a "$x$" instead of a "$y$", probably because it was too tiny too be noticed. I fixed that and made the fractions bigger. =) – user21820 Apr 21 '17 at 15:13
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Their proof is kind of crappy. Just take the log of both sides of the equation, and you'll get log(x)log(y)=log(y)log(x)

Jerry Guern
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For an intuitive (purposefully non-rigorous) reason, consider writing logarithms in an infix notation, as $a \underline{\log} b$ rather than $\log_b a$.

Then the relation is

$$x \text{ ^ } (y \underline{\log} b) = y \text{ ^ } (x \underline{\log} b)$$

which is just one arithmetic step up from a relation you probably already know:

$$x \times (y \div b) = y \times (x \div b)$$

which itself is also just one arithmetic step up from:

$$x + (y - b) = y + (x - b)$$

It makes a nice pattern.

DanielV
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  • Hello! I strongly urge you to either delete this answer or to explain in full the analogy. The reason is that (as you definitely know) one cannot do mathematics by following intuitive patterns. Worse still, in this case it isn't at all obvious what the pattern is, and exponentiation is not even commutative. There is a deep algebraic reason (group actions and their inverses), but maybe not one that should be gotten into at this level. Agree? =) – user21820 Apr 21 '17 at 15:28
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Method-1: Notice, we know that $\large \color{red}{\frac{1}{\log_m(n)}=\log_n(m)}$ & $\color{red}{m^{\log_m(n)}=n}$

Now, we have $$\large x^{\log_b(y)}=y^{\log_b(x)}$$ $$\large \iff x^{\frac{1}{\log_b(x)}}=y^{\frac{1}{\log_b(y)}}$$ $$\large \iff x^{\log_x(b)}=y^{\log_y(b)}$$ $$\large \iff b=b$$ Method-2: we have $$\large x^{\log_b(y)}=y^{\log_b(x)}$$ taking log with base $x$ on both the sides, we get
$$\large \log_x(x^{\log_b(y)})=\log_x(y^{\log_b(x)})$$ $$\log_b(y)\log_x(x)=\log_b(x)\log_x(y)$$ $$\frac{\log_b(y)}{\log_b(x)}=\log_x(y)$$ $$\log_b(y)\cdot \log_x(b)=\log_x(y)$$ $$\log_x(y)=\log_x(y)$$

Harish Chandra Rajpoot
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