We start with:
#5^(2xx x)-5^(x+3)+125=5^x#
Let's first see that #125=5^3#:
#5^(2xx x)-5^(x+3)+5^3=5^x#
We can use the rules #x^a xx x^b=x^(a+b)# and #(x^a)^b=x^(ab)# to untangle the expressions:
#(5^x)^2-(5^3)5^x+5^3=5^x#
Let's try subtracting #5^x# from both sides to get the #x# terms all on the left:
#(5^x)^2-(5^3)5^x+5^3-5^x=0#
We can combine the #5^x# terms and see that we'll have #-5^3-1=-125-1=126# of them:
#(5^x)^2-(124)5^x+5^3=0#
Let's set #a=5^x#:
#a^2-126a+125=0#
We can now factor this:
#(a-125)(a-1)=0#
#a=1, 125#
Let's now substitute back in:
#5^x=1, 125#
And take each solution separately:
#5^x=1=>x=0#
#5^x=125=5^3=>x=3#
Let's check the answers:
#5^(2xx x)-5^(x+3)+125=5^0#
#5^(2xx 0)-5^(0+3)+125=1#
#5^0-5^3+5^3=1#
#1=1color(white)(000)color(green)sqrt#
~~~~~
#5^(2xx x)-5^(x+3)+125=5^x#
#5^(2xx 3)-5^(3+3)+125=5^3#
#5^6-5^6+125=5^3#
#125=125 color(white)(000)color(green)sqrt#