Question

$3^{x} = 5^{2x} ; 3^{x+3}+ 3^{x} = 7^{x+1}+5* 7^{x} ; 2^{8-x}+ 7^{3-x}= 7^{4-x} + 2^{3-x}*11$

Answer 1

1
3^x=25^x/25^x
(3/25)^x=1
x=0
2
3^x*(27+1)=7^x*(7+5)
28*3^x=12*7^x
(3/7)^x=12/28
(3/7)^x=3/7
x=1
3
2^8/2^x+7³/7^x=7^4/7^x+2³*11/2^x
2^8*7^x+7³*2^x=7^4*2^x+2³*11*7^x
2^8*7^x-2^3*11*7^x=7^4*2^x-7^3*2^x
2³*7^x*(32-11)=7³*2^x*(7-1)
2³*21*7^x=7³*6*2^x
(7/2)^x=7³**2*3/(2³*3*7)
(7/2)^x=(7/2)²
x=2