Question

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

Answer 1

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`