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.
