Asked by kayla
how do i solve this using logarithms:
1) 4^(x)=8
2) 2^(x) * 8^(-x)= 4^(x)
1) 4^(x)=8
2) 2^(x) * 8^(-x)= 4^(x)
Answers
Answered by
drwls
1) x log 4 = log 8
x = log 8/log 4 = 1.500 = 3/2
(any base can be used)
2. 2^x * 8^(-x) = 4^x
2^x *(2^3)^-x = (2^2)^x
2^x *(2^-3x) = 2^2x
2^-2x = 2^2x
-2x = 2x
x = 0
If you want to use logs,
x log 2 -x log 8 = x log 4
x *(log2 - log 8)
= x log (1/4) = x log 4
x*(log (1/4) - log 4) = 0
x*(-2 log 4) = 0
Therefore x = 0
x = log 8/log 4 = 1.500 = 3/2
(any base can be used)
2. 2^x * 8^(-x) = 4^x
2^x *(2^3)^-x = (2^2)^x
2^x *(2^-3x) = 2^2x
2^-2x = 2^2x
-2x = 2x
x = 0
If you want to use logs,
x log 2 -x log 8 = x log 4
x *(log2 - log 8)
= x log (1/4) = x log 4
x*(log (1/4) - log 4) = 0
x*(-2 log 4) = 0
Therefore x = 0