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

$${f}^{'}(x)={(\frac{3}{5})}^{x}\mathrm{ln}(\frac{3}{5})+{(\frac{4}{5})}^{x}\mathrm{ln}(\frac{4}{5})<0 \forall x\in R$$f′(x)=(53​)xln(53​)+(54​)xln(54​)<0∀x∈R

$${\because \frac{d}{dx}{a}^{x}={a}^{x}\mathrm{ln}a}$$∵dxd​ax=axlna

Hence, $f(x)$f(x) is monotonically decreasing.

Also, $\underset{x\rightarrow +\infty }{\mathrm{lim}}f(x)\rightarrow -1$x→+∞lim​f(x)→−1 ${\because \underset{x\rightarrow \infty }{\mathrm{lim}}{a}^{x}=0,0<a<1}$∵x→∞lim​ax=0,0<a<1

and $\underset{x\rightarrow -\infty }{\mathrm{lim}}f(x)\rightarrow \infty$x→−∞lim​f(x)→∞ ${\because \underset{x\rightarrow -\infty }{\mathrm{lim}}{a}^{x}=\infty ,0<a<1}$∵x→−∞lim​ax=∞,0<a<1

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