# Solve

#### eulid

Solve for $$\displaystyle x$$

$$\displaystyle 5^\sqrt{x}+12^\sqrt{x}=13^\sqrt{x}$$

Answer is $$\displaystyle x=4$$. How to solve it? By taking the logarithms of both sides? Please give some hints. Thank you.

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#### skipjack

Forum Staff
The equation can't be solved algebraically. It may be possible to use calculus to show that it has only one solution. It would then follow that $x = 4$ is the only solution.

1 person

#### topsquark

Math Team
Solve for $$\displaystyle x$$

$$\displaystyle 5^\sqrt{x}+12^\sqrt{x}=13^\sqrt{x}$$

Answer is $$\displaystyle x=4$$. How to solve it? By taking the logarithms of both sides? Please give some hints. Thank you.
Aside from recognizing that (5, 12, 13) is a Pythagorean triple the only thing I can think of it to graph $$\displaystyle y = 5^{ \sqrt{x} } + 12^{ \sqrt{x} } - 13^{ \sqrt{x} }$$.

You can't actually prove that x = 4 this way (or prove that it's the only solution), but it will give you the idea and you can put it into the equation and find out.

-Dan

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#### DarnItJimImAnEngineer

Graphing would have been my method of choice, as well.

Interesting side note, while I believe $x=4$ is the only actual solution, if you account for complex numbers, I think there are an infinite number of values of $x$ that come arbitrarily close to solving the equation.

Let $f(x) = 5^{\sqrt{x}}+12^{\sqrt{x}}-13^{\sqrt{x}}$
If we let $t \equiv \sqrt{-x}$, then
$f(x) = g(t) = 5^{it} + 12^{it} - 13^{it} = e^{\ln(5) it} + e^{\ln(12) it} - e^{\ln(13) it}$,
or the sum of three sinusoidal waves at frequencies (in t) of ln(5), ln(12), and ln(13).

Since these frequencies are irrational, I don't think the phases will ever exactly synch up except at $x=t=0$, so the magnitude will never be identically zero. It will come arbitrarily close over and over again, though.

For example, $|f(-28854)| = 0.0149$. At more negative numbers, the local minima start getting smaller. Unfortunately, the "beat" frequency is so low that you have to go to really large negative values of $x$ to find significantly smaller magnitudes of f. They are more or less predictable, however, so we could find them were we so inclined.

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#### idontknow

If $$\displaystyle \sqrt{x}$$ is an integer , apply modular arithmetics.

#### romsek

Math Team
You solve this as Topsquark noted by seeing it's a Pythagorean triple.

$5^2 + 12^2 = 13^2$

$\sqrt{x} = 2$

$x=4$

#### topsquark

Math Team
If $$\displaystyle \sqrt{x}$$ is an integer , apply modular arithmetics.
I'm not following you. What do you mean?

-Dan

#### Yooklid

Doesn't Fermat's last theorem require that $\sqrt x$ must be less than 3? That would limit the number of possible values. Of course you have to assume Fermat's last theorem is valid (or prove Fermat's last theorem).

#### topsquark

Math Team
Doesn't Fermat's last theorem require that $\sqrt x$ must be less than 3? That would limit the number of possible values. Of course you have to assume Fermat's last theorem is valid (or prove Fermat's last theorem).
It's a good point, but no one specificed that x is an integer. FLT only holds for integer powers.

-Dan

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#### Yooklid

That's the problem when these kinds of problems are posed. There's no context. So we have to be able to read the mind of whichever school teacher or university professor came up with it. Does it have to be an integer solution, or can it be reals or complex? Can x be negative? I question why they bothered to include a radical around the x. Since x never appears without the radical, the radical is meaningless.