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

Question

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!

Answer 1

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.

Answer 2

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.

Answer 3

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 $({1\over 3})^y+({1\over 4})^y>({1\over3})^y > ({1\over7})^y$

Answer 4

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 $m<n$

Answer 5

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.