exponentail thingies
easy, look at all them them, see that they have 5 in common?
rremember how esay it was to factor
ax^2+bx+c=0
now we have
5^(2x)-6(5^x)+5=0
remember that 5^(2x)=(5^2)^x or (5^x)^2
in other words, we can rewrite it as
1(5^x)^2-6(5^x)+5=0
if yo want, replace 5^x with a and factor
1a^2-6a+5=0
(a-1)(a-5)=0
a=5^x
(5^x-1)(5^x-5)=0
set each to zero
5^x-1=0
5^x=1
take the log₅ of both sides
x=log₅1
5^x-4=0
5^x=4
take the log₅ of both sides
x=log₅4
x=log₅1 and/or log₅4
second quesiton
same thing
1(2^x)-10(2^x)+16=0
factor
(2^x-8)(2^x-2)=0
set each to zero
2^x-8=9
2^x=8
x=3
2^x-2=0
2^x=2
x=1
x=3 or 1
first one
x=log₅1 and/or log₅4
second one
x=1 and/or 3
