How do you solve -\log 8= 3- \log x?

1 Answer
May 13, 2017

Isolate logx then use the definition of a logarithm to solve for x. Answer: 8000

Explanation:

Question: -log8=3-logx

Note that log(x) is the same as log_(10)(x)

We can see that -log8 and 3 are both constants, so we move them to one side and make logx positive:
logx=3+log8

Now we can use the definition of a logarithm (log_(b)a=c -> b^c=a) to take x out of the log:
x=10^(3+log8)

=10^3*10^(log8)

Since, exponential and logarithmic are inverse functions, they cancel each other:
=1000*8

=8000