Question

Find the solution set of $f^{\prime }\left(x\right)>g^{\prime }\left(x\right)$ where $f\left(x\right)=\left(\frac{1}{2}\right)5^{2x+1}$ and $g\left(x\right)=5^x+4x\log 5.$

• A. $(1,\infty )$
• B. $(0,1)$
• C. $[0,\infty )$
• D. $(0,\infty )$

Answer 1

The correct option is D $(0,\infty )$

As $f\left(x\right)=\left(\frac{1}{2}\right)5^{2x+1}$

$\Rightarrow f^{\prime }\left(x\right)=\left(\frac{1}{2}\right)5^{2x+1}\left(\log 5\right)\times 2=\left(\log 5\right).5^{2x+1}$

And $g\left(x\right)=5^x+4x\log 5\Rightarrow g^{\prime }\left(x\right)=5x\log 5+4\log 5=\left(\log 5\right)(5x+4)$

Solving $f^{\prime }\left(x\right)>g^{\prime }\left(x\right)$

$\Rightarrow \left(\log 5\right).5^{2x+1}>\left(\log 5\right).(5x+4)\Rightarrow 5^{2x+1}>5^x+4$

Substitute $5^x=t,$ we get

$5^{2x}.5>5^x+4\Rightarrow (5^x{)}^2.5>5^x+4$

$\Rightarrow 5t^2+t-4>0\Rightarrow (5t+4)(t-1)>0\Rightarrow t>1$ or $t<-\frac{4}{5}$

Because since $t=5^x$

$\therefore t>1\Rightarrow x=(0,\infty )$