Answer to Question #101407 in Algebra for Israel

Question

1) 3^2x-4(3^x-2)+1=0
2) 125^x²-0.2(25^x)²=0

Answer 1

1)

$3^{2x}-4(3^x-2)+1=0,$

substitution $t=3^x,$ $t^2=3^{2x},$

t²-4(t-2)+1=0

t²-4t+9=0

D=16-36= -20<0,

there is no real solution.

2)

$125^{x^2}-0.2(25^x)^2=0$

$125^{x^2}=5^{-1}25^{2x}$

$5^{3x^2}=5^{4x}5^{-1}$

$3x^2=4x-1$

3x²-4x+1=0

D=16-12=4

x=(4-2)/(2*3) and x=(4+2)/(2*3),

x=1/3 and x=1.