PreCalculus
posted by Caroline on .
I'm having problems understanding how to do these problems.
25. 32^(2x3)=2
27. 1/4= 2^3x
29. 9^x=27
33. 81^(x1)= 27^2x

recall that log(x^a) = a * log x
25)
32^(2x3) = 2
take log of both sides to get
(2x3) log 32 = log 2
2x3 = log 2 / log 32
Now recall that loga/logb = log_b(a)
so, log2/log32 = log_32(2) = 1/5
since 2^5 = 32, 2 = 32^(1/5), so log_32(2) = 1/5
2x3 = 1/5
2x = 16/5
x = 8/5
Or, we could have started out by noting that 2 = 32^(1/5), so equating powers of 32,
2x3 = 1/5
...
27)
similarly, since 1/4 = 2^2,
2 = 3x
x = 2/3
29)
Since 3= 9^1/2, 27 = 3^3 = 9^3/2, so
x = 3/2
33)
27 = 3^3
81 = 3^4, so
27 = 81^(3/4)
equating powers,
x1 = 2x(3/4)
x1 = 3x/2
x/2 = 1
x = 2
check:
81^3 = 3^12
27^6 = 3^12