Question

Find the number of solutions of equation $2^x+3^x+4^x-5^x=0$

Answer 1

We have, $2^x+3^x+4^x=5^x$
Divide both sides by $5^x$
${\left(\frac{2}{5}\right)}^x+{\left(\frac{3}{5}\right)}^x+{\left(\frac{4}{5}\right)}^x=1$
Let $f\left(x\right)={\left(\frac{2}{5}\right)}^x+{\left(\frac{3}{5}\right)}^x+{\left(\frac{4}{5}\right)}^x$
Now $f^{\prime }\left(x\right)={\left(\frac{2}{5}\right)}^x\ln \frac{2}{5}+{\left(\frac{3}{5}\right)}^x\ln \frac{3}{5}+{\left(\frac{4}{5}\right)}^x\ln \frac{4}{5}$
Clearly $f\left(x\right)<\forall x\in R$ since $\ln a<0$ for $0<a<1$
Hence f is strictly decreasing function
Now as $x\to -\infty \Rightarrow f\left(x\right)\to \infty$
And as $x\to \infty \Rightarrow f\left(x\right)\to 0$
Hence range of range of f is $(0,\infty )$
Hence there will be exactly one roots of $f\left(x\right)=1$