Question 662425
  <pre><font face = "Tohoma" size = 3 color = "indigo"><b> 
Hi,previously Posted
1. 5^2x-6(5^x)+5=0
 (5x)^2-6(5^x)+5=0
 let {{{5x=y }}}    || Good work!
 y^2-6y+5=0
factoring
(y-5)(y-1) = 0,  y = 5 0r y = 1  AND   x = 1  0r x = 1/5
As to the Remainder: the Idea is combining Like Terms...
2.4^x(5^2x)=10  &#8658; {{{2^(2x)5^(2x) = 10^(2x) = 10}}}, x = 1/2 
 3.7^x+1-2=2(7^x)+3 &#8658; 7^x(7-2) = 5 &#8658; 7^x = 1,   x = 0 
 4.{{{6(3^(x-1))=3^4-3^x}}} &#8658; 
   3^x (6*3^(-1)+1) = 3^x(2+1)= 81  &#8658;  3^x = 27,    x = 3