Question

Find the greatest value of the function $y=7+2x\ln 25-5^{x-1}-5^{2-x}.$

Answer 1

$y=7+2x\ln 25-5^{x-1}-5^{2-x}$

$y^{\prime }=0+2\ln 5^2-5^{x-1}\ln 5+5^{2-x}\ln 5$.

$\left[\frac{d}{dx}\right(a^x)=a^x\ln a]$

$y^{\prime }=4\ln 5-5^{x-1}\ln 5+\frac{\ln 5}{5^{x-2}}=0$

$=(4\ln 5-5^{x-1}\ln 5)5^{x-2}+\ln 5=0$

$=\ln 5\left(\right[4\times 5^{x-2}-5^{(x-2)+(x-1)}]+1)=0$

$=\ln 5(4\times 5^{x-2}-5^{2x-3}+1)=0$

$\therefore 4\times 5^{x-2}-5^{2x-3}+1=0$

when $x=2$,

$4-5+1=0$

Now, $y^{\prime \prime }=0-5^{x-1}(\log 5{)}^2+(\ln 5{)}^2{5}^{-x+2}(-1)$

Put $x=2$

$y^{\prime \prime }=-(\log 5{)}^2\times 5-(\ln 5{)}^2<0$

$\therefore x=2$ is the point of maxima.

Put $x=2$ in $y=7+2x\ln 25-5^{x-1}-5^{2-x}$

$y=7+8\ln 5-5-1$

$=1+8\ln 5$.