How do you solve #log x=-2#?

1 Answer
Jun 19, 2016

Answer:

#x=1/100#.

Explanation:

You only need to know that the logarithm is the inverse function of the exponential.

If by "#log#" you mean the base 10 logarithm, then simply make both terms the exponents of a #10#-based power:

#log(x)=-2 \implies 10^{log(x)}=10^{-2}#

Now, as I said, exponential and log are inverse function, which means that they cancel out, remaining with

#x=10^{-2}=1/100#

If you have another base, simply substitute that, replacing 10 with your base.