COB and COD complementary
AOB and BOC adjacent angle that is neither comp. or supp.
AOC and COD supplementary
AOE and COD vertical
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
