hipa said:you can write the equation as (5^x)^2-(5^x)(5)+4=0
Now substitute t=5^x. Now we get t^2-5t+4=0. This is a simple quadratic in t and can be factored as (t-4)(t-1)=0. Which gives t=4 or t=1. Now put back 5^x=t. You get 5^x=4 or 5^x=1, which means x=log_5(4) or x=0. Hope this helps.
ahh yes yes. i was about to say thathipa said:you can write the equation as (5^x)^2-(5^x)(5)+4=0
Now substitute t=5^x. Now we get t^2-5t+4=0. This is a simple quadratic in t and can be factored as (t-4)(t-1)=0. Which gives t=4 or t=1. Now put back 5^x=t. You get 5^x=4 or 5^x=1, which means x=log_5(4) or x=0. Hope this helps.

hipa said:you can write the equation as (5^x)^2-(5^x)(5)+4=0
Now substitute t=5^x. Now we get t^2-5t+4=0. This is a simple quadratic in t and can be factored as (t-4)(t-1)=0. Which gives t=4 or t=1. Now put back 5^x=t. You get 5^x=4 or 5^x=1, which means x=log_5(4) or x=0. Hope this helps.