Question

5^2(x-1)+5^x+1=0.04

25

Answer 1

$5^{2(x-1)}+5^{x+1}=\frac{4}{100}=\frac{1}{5^2}$

$5^{2x}*5^{-2}+5^x*5=5^{-2}$

multiplying both side by $5^{2}$ , we get

$5^{2x}+5^x*5^{2+1}=1$ ($a^p*a^q=a^{p+q}$ )

$5^x*5^x+5^x*5^3=1$ ; on dividing the equation by 5^x in both side, we get

$5^x+5^3=\frac{1}{5^x}$ or $5^x-\frac{1}{5^x}=-125$

let $5^x=y$ , then the above equation can be written as,

$y-\frac{1}{y}=-125$ or $y^2+125y-1=0$

$y=\frac{-125\pm\sqrt{125^2-4*1*(-1)}}{2*1}$ $=\frac{-125\pm(125.015999)}{2}$

y= -125.008 or y= 0.0079995

we know that $y=5^x$ , for any value of x, gives the value of y always positive. so

y = 0.0079995 is only the valid value.

so, $y=0.0079995=5^x$

taking log (base 10) both side of equation, we get

$log(0.0079995)=log(5^x)$

$-2.09693716 = x * log(5)$ $(log(a^p)=p*log(a))$

$x=\frac{-2.09693716}{0.698970004}$

$x=-3.00003884$