If f x = 3 5 x + 4 5 x - 1 ,   x ∈ R , then the equation f x = 0 has :

$f^{\text{'}}\left(x\right)={\left(\frac{3}{5}\right)}^x\ln \left(\frac{3}{5}\right)+{\left(\frac{4}{5}\right)}^x\ln \left(\frac{4}{5}\right)<0\mathrm{ }\forall \mathrm{ }\mathrm{x}\in \mathrm{R}$

$\left\{\because \frac{d}{dx}{a}^x=a^x\ln a\right\}$

Hence, $f\left(x\right)$ is monotonically decreasing.

Also, $\underset{x\to +\infty }{\lim }f\left(x\right)\to -1$ $\left\{\because \underset{x\to \infty }{\lim }{a}^x=0, 0<a<1\right\}$

and $\underset{x\to -\infty }{\lim }f\left(x\right)\to \infty$   $\left\{\because \underset{x\to -\infty }{\lim }{a}^x=\infty , 0<a<1\right\}$

and there is only $1$ solution for $f\left(x\right)=0$.