Is this the correct procedure for this kind of question? Apparently the answer is supposed to be log6/log5.
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Answer:
log 3
---------- = x
log 5
Step-by-step explanation:
25 ^x - 5^x +4 = 10
Subtract 4 from each side
25 ^x - 5^x +4-4 = 10-4
25 ^x - 5^x = 6
25 can we rewritten as 5^2
5^2^x - 5^x = 6
a^b^c can be written as a^(b*c)
5^(2x) - 5^x = 6
Replace 5^x with m
m^2 - m = 6
Subtract 6 from each side
m^2 -m -6 = 6-6
m^2 -m-6=0
Factor
(m-3) (m+2) = 0
Using the zero product property
m-3 = 0 m+2=0
m=3 m=-2
Substitute m=x^2 back in
5^x =3 5^x =-2
Take the log of each side
log5 (3) =x log5 (-2) =x
There is no log -2 so this is not a solution
Assuming we want to use base 10
log 3
---------- = x
log 5