# Pre-Calculus

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I'm having problems understanding how to do these problems.

25. 32^(2x-3)=2

27. 1/4= 2^3x

29. 9^x=27

33. 81^(x-1)= 27^2x

• Pre-Calculus -

recall that log(x^a) = a * log x

25)
32^(2x-3) = 2
take log of both sides to get
(2x-3) log 32 = log 2
2x-3 = 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

2x-3 = 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,
2x-3 = 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,
x-1 = 2x(3/4)
x-1 = 3x/2
x/2 = -1
x = -2
check:
81^-3 = 3^-12
27^-6 = 3^-12

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