Question

If $v\left(x\right)$ is larger of $e^x-1$ and $(1+x)\log (1+x)$ for $xϵ(0,\infty )$ then $log\left(v\right(8)+1)$ is equal to

Answer 1

$atx=0bothfuctionsareequalandderivateof{e}^x-1=\frac{d(e^x-1)}{dx}=e^x>0derivateof(1+x)\log (1+x)=\frac{d\left(\right(1+x\left)\log (1+x)\right)}{dx}$

$=(1+x)\frac{d\left(\log (1+x)\right)}{dx}+\log (1+x)\frac{d(1+x)}{dx}=(1+x).\frac{1}{(1+x)}+\log (1+x)=1+\log (1+x)>0andforx>0e^{x}isgreaterthan1+\log (1+x).thatmeansthat{e}^x-1>(1+x)\log (1+x)soV\left(x\right)=e^x-1V\left(8\right)=e^8-1V\left(8\right)+1=e^8\log \left(V\right(8)+1)=8.$