Question

How do you solve `log x - log 10= 14`?

Answer 1

`x=10^15`

Explanation:

We make use of two identities, one - `loga-logb=log(a/b)` and if `logu=v`, then `u=10^v`

Using them we can write `logx-log10=14` as

`log(x/10)=14`

or `x/10=10^14`

i.e. `x=10xx10^14=10^15`