SOLUTION: Pls help me evaluate this Log32(0.0625)

Question 758487: Pls help me evaluate this Log32(0.0625)
Found 2 solutions by Theo, MathTherapy:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
your expression is:
log32(.0625)
let y = log32(.0625)
this is true if and only if 32^y = .0625
take the log of both sides of this equation to the base of 10 to get:
log10(32^y) = log10(.0625)
this becomes y * log10(32) = log10(.0625)
divide both sides of this equation by log10(32) to get:
y = log10(.0625) / log10(32)
since log10 is the LOG function of your calculator, you can use your calculator to find:
y = LOG(.0625) / LOG(32) to get:
y = -.8
32^-.8 = .0625 so it looks like y = -.8 is your answer.
your answer is:
log32(.0625) = -.8

the general log properties that are employed in the solving of this problem are:

y = logb(x) if and only if b^y = x
logb(x^a) = a * logb(x)
if x = y, then logb(x) = logb(y)

logb means log to the base of b.
logb(x) means log of x to the base of b.

Answer by MathTherapy(10555) (Show Source): You can put this solution on YOUR website!
Pls help me evaluate this Log32(0.0625)

You can simply divide, as follows: to get

You can simply enter it into a calculator as log(0.0625), with a base of 32. This will also give you:

------ Logarithmic form

By letting the exponent be E, we get:
------ Exponential form

----- Converting 0.0625

----- Converting 32

5E = - 4 ------- Setting exponents equal since bases are equal

E, or exponent = , or
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