Question

What are possible values of x if `2logx<log(2x-1)?`?

Answer 1

No possible solutions.

Explanation:

First, it is always a good idea to identify the domain of your logarithm expressions.

For `log x`: the domain is `x > 0`
For `log(2x-1)`: the domain is `2x - 1 > 0 <=> x > 1/2`

This means that we only need to consider `x` values where `x > 1/2` (the intersection of the two domains) since otherwise, at least one of the two logarithm expressions is not defined.

Next step: use the logarithm rule `log(a^b) = b * log(a)` and transform the left expression:
`2 log(x) = log(x^2)`

Now, I'm assuming that the basis of your logarithms is `e` or `10` or a different basis `>1`. (Otherwise, the solution would be quite different).

If this is the case, `log(f(x)) < log(g(x)) <=> f(x) < g(x)` holds.

In your case:
`log(x^2) < log(2x - 1)`
`<=> x^2 < 2x - 1`
`<=> x^2 - 2 x + 1 < 0`
`<=> (x-1)^2 < 0`

Now, this is a false statement for all real numbers `x` since a quadratic expression is always `>=0`.

This means that (under the assumption that your logarithm basis is indeed `>1`) your inequality has no solutions.