Question

find solution from exponent and powers chapter​

Answer 1

Solution:

2^x2: 2^2x=8:1

2^x2/2^2x=8/1

2^x2-2x=2^3

By comparing,

x^2-2x-3=0

x^2-3x+x-3=0

x(x-3)+1(x-3)=0

(x-3)(x+1)=0

So, x=3 or -1

Answer 2

$\mathrm{ 2^{\normal x^2}:2^{2x}=8:1}\\\\\\\mathrm{\dfrac{2^{x^2}}{2^{2x}}=8}\\\\\\\mathrm{2^{x^2-2x}=2^3}\\\\\implies x^2-2x=3\\\\x^2-2x-3=0\\\\x^2-(x-3x)-3=0\\\\x^2-x+3x-3=0\\\\x(x-1)+3(x-1)=0\\\\(x+3)(x-1)=0\\\\x=-3,1$