# Solve equations like $3^x+4^x=7^x$

How can I solve something like this?

$$3^x+4^x=7^x$$

I know that $x=1$, but I don't know how to find it. Thank you!

• There is no elementary method to compute solutions of such equations. For example, there is no closed form for the solution of $3^x+4^x=8^x$ – Crostul Nov 6 '15 at 8:55
• Then how can I solve this one? – Victor Nov 6 '15 at 8:56
• as an intuitive proof, it may help to graph these functions – Elliot G Nov 6 '15 at 8:56
• @Victor The only thing one can do to solve it is to guess the solution. Obviously, in this case one notes that $3+4=7$, so plugging $x=1$ in your equation you get a correct identity. But for the general case, only numerical methods work. – Crostul Nov 6 '15 at 8:58
• What do you mean "how to find it"? $3^1+4^1=7^1$, thus it is a solution to the equation. The hard part is to show this is the only solution (if it is). – Ove Ahlman Nov 6 '15 at 8:58

## 5 Answers

Here $\displaystyle 3^x+4^x = 7^x\Rightarrow \bf{\underbrace{\left(\frac{3}{6}\right)^x+\left(\frac{4}{6}\right)^x}_{Strictly\ decreasing\; function}} = \underbrace{\left(\frac{7}{6}\right)^x}_{Strictly\; increasing\; function}$

So these two curve Intersect each other exactly one Point.

So we can easily calculate $x=1$ is only Solution of above equation.

• Woah, good job! – Paolo Leonetti Nov 6 '15 at 9:40
• Why $\frac{3}{6}$? Why $6$? – Victor Nov 8 '15 at 10:40
• Why not 6? If you don't like 6 use 5. or 4.98735463... or ... 4 ..or 7. The point is max(3,4) <= 6 <= 7 so by putting these over 6 you get the LHS is decreasing and the RHS is increasing (or constant if you used 7). So as $3^0 + 4^0 > 7^0$ there has to be exactly one solution. In fact we can use this argument to prove that if a,b < c then $a^x + b^x = c^x$ has exactly one solution. Finding out what that solution is may be hard as hell though. In this case, however, it was easy as sin. – fleablood Jun 22 '16 at 17:59
• Hmm... I guess the one thing this very elegant answer didn't consider is that we have to show the RHS is increasing and unbounded and the LHS is decreasing with a limit to 0. If the RHS is increasing but assymptotic to a value less than the limit of the LHS it'd be possible they never intersect. ... oh wait. That'd be zero solutions and we know $x = 1$ is a solution.... never mind. – fleablood Jun 22 '16 at 18:08

For $x>1$ obviously $(3+4)^x > 3^x+4^x$ by binomial theorem.

For $1>x>0$, we have $(3^x+4^x)^{1\over x} > 3^{x{1\over x}}+4^{x{1\over x}}$ since ${1\over x} > 1$ and hence $3^x+4^x > (3+4)^x$

For $x< 0$, let $y=-x$ then $({1\over 3})^y+({1\over 4})^y>({1\over3})^y > ({1\over7})^y$

HINT:

As for $0<a<1,$ $$a^m>a^n$$ if $m<n,$

$$\left(\dfrac37\right)^m+\left(\dfrac47\right)^m>\left(\dfrac37\right)^n+\left(\dfrac47\right)^n$$

if $m<n$

Defining $f(x):=\mathrm{log}_7\left(3^x+4^x\right)$, you want to search for fixed points of $f$. But $$f^\prime(x)=\frac{1}{\ln 7}\cdot \frac{3^x\ln 3+4^x \ln 4}{3^x+4^x}>0$$ and $$f^{\prime\prime}(x)=\frac{1}{\ln 7}\cdot \frac{3^x4^x}{(3^x+4^x)^2} \cdot ((\ln 4)^2+(\ln 3)^2-2\ln 3 \ln 4)$$ which is positive too by arithmetic-geometric mean inequality.

By putting the given equation in the form:

$${\left( \dfrac {3}{3+4} \right ) } ^x + {\left( \dfrac {4}{3+4} \right ) } ^x =1$$

we find it is satisfied by $x=1$. The monotonic nature of exp function gives no other real roots.