Exponential Rational Equation?

Relevant wiki: General Diophantine Equations - Problem Solving

Let $x = \frac{p}{q}$ where $p,q$ are coprime integers and $q$ is positive. Then the equation becomes

$\displaystyle \left( \frac{p}{q} \right) ^ {\frac{2p-q}{q}} = 2$

Raising each side to the power of $q$ gives

$\displaystyle \left( \frac{p}{q} \right) ^ {2p-q} = 2^q$

Multiplying both sides by $q^{2p-q}$ (which is nonzero because $q$ is positive) gives

$p^{2p-q} = q^{2p-q} \cdot 2^q$

We now divide into three cases.

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Case 1 : $2p-q = 0$

This is impossible, since the equation becomes $1 = 1 \cdot 2^q$ , or $q = 0$ , contradicting that $q$ is positive.

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Case 2 : $2p-q > 0$

Now both sides are integers. The right hand side is divisible by $q^{2p-q}$ , so the left hand side must also be divisible by it. Thus $q^{2p-q} | p^{2p-q}$ , or $q | p$ . But since $p,q$ are coprime, we must have $q = 1$ , reducing to $p^{2p-1} = 2$ .

Now, note that if $p \ge 2$ , then $p^{2p-1} \ge 2^3 = 8 > 2$ , so there's no solution there. If $p = 1$ , then $p^{2p-1} = 1^1 = 1 \neq 2$ , so it's not a solution. If $p \le 0$ , then $2p-q < -1$ , contradicting our assumption of the case. So this case doesn't give any solution.

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Case 3 : $2p-q < 0$

Multiply both sides by $(pq)^{q-2p}$ . We now have $q^{q-2p} = p^{q-2p} \cdot 2^q$ , and both sides are now integers. Similar to Case 2, we have $p | q$ , so $p = 1$ and the equation becomes $q^{q-2} = 2^q$ . Remember that $q$ is positive.

Now, if $q$ has an odd prime factor $k$ , then $q \ge 3$ , so $q-2 \ge 1$ . Thus the left hand side $q^{q-2}$ is divisible by $q$ (because the exponent is positive), which in turn is divisible by $k$ (because it's a divisor of $q$ ). So the left hand side is divisible by $k$ . But the right hand side isn't, since it's a power of 2. Thus $q$ doesn't have any odd prime factor, and so $q = 2^k$ for some $k$ . The equation reduces to $2^{k(q-2)} = 2^q$ , or $k(q-2) = q$ .

Now, we can check that $q = 1, 2$ doesn't give a solution ( $q = 1$ leads to $k = 0$ , while $q = 2$ leads to $q-2 = 0$ , so the left hand side is 0, but the right hand side isn't). Otherwise, $q \ge 4$ ( $q = 3$ is not a power of 2) and so $q-2 \ge 2$ , so $k = \frac{q}{q-2} = 1 + \frac{2}{q-2} \le 1 + \frac{2}{2} = 2$ . Thus the only candidate remaining is $k = 2$ or $q = 4$ , which is indeed a solution.

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Thus the only solution is $x = \frac{p}{q} = \boxed{\frac{1}{4}}$ , which can be verified to work.

Writing $x = 2^{n-1}$ we get $2^n = \frac n{n-1}.$ Since $n$ and $n-1$ are necessarily coprime this fraction cannot be reduced any further.

If $n$ is not rational, then $2x = 2^n = n/(n-1)$ cannot be rational, either, contrary to the assumption.

If $n$ is rational but not an integer, then $2^n$ cannot be rational, but since $2^n = n/(n-1)$ it must be rational; a contradiction.

Thus it follows that $n$ is an integer.

If $n$ is positive, then $2^n$ is an integer, so that in the fraction $n/(n-1)$ the denominator is equal to $\pm 1$ . This requires $n = 0$ or $n = 2$ , and it is easy to see that these are not solutions.

If $n$ is negative, then $2^n = 1/K$ with $K$ and integer, so that in the fraction $n/(n-1)$ the numerator is equal to $\pm 1$ . This requires $n = 1$ (which would cause division by zero) or $n = -1$ . The latter is indeed a solution, since $2^{-1} = \frac{-1}{(-1)-1} = \frac 12;$ it follows that $x = 2^{n-1} = 2^{-1-1} = \frac 14 = \boxed{0.25}.$

X^(2x-1) = 2 X^(2x) = 2x Let a = 2x (.5a)^a = a (.5)^a times (a)^a = a (.5)^a times a^(a-1) = 1 a^(a-1) = 2^a This means that a has to be a power of 2 Try simple powers of 2 to find that a = 1/2 So X = 1/4

$x^{2x - 1} = 2 \iff x^{2x} = 2x \iff 2x = \frac{\log 2x}{\log x} \iff 2x = 1 + \frac{\log 2}{\log x}$ If $x$ is a rational number, so is the right hand side in this equation, and thus so is $\frac{\log 2}{\log x}$ . If $\frac{\log 2}{\log x} = \frac{p}{q}$ then $\log 2 = \frac{p}{q} \log x \implies \log 2 = \log x^{\frac{p}{q}} \implies 2 = x^{\frac{p}{q}} \implies 2^q = x^p$ . If $x$ is a rational number and $p,q$ are integers, the only way for this equation to hold true is if $x = 2^b$ where $b$ must be an integer. Substituing $x = 2^b$ in $2x = 1 + \frac{\log 2}{\log x}$ yields $2^{b+1} = 1 + \frac{1}{b}$ .

If $b$ was a positive integer, the left-hand side is too big compared to the right-hand side, and if we let $b$ be negative, the left-hand side is able to go to 0 at an exponential speed, so any solutions will have small absolute value. Testing small values we find that $b = -2$ is a solution yielding $x = 2^{-2} = \frac{1}{4}$ .

An algebraical solution is wanted.