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

Nov 9, 2015

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 $\log \left(2 x - 1\right)$: the domain is $2 x - 1 > 0 \iff x > \frac{1}{2}$

This means that we only need to consider $x$ values where $x > \frac{1}{2}$ (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 $\log \left({a}^{b}\right) = b \cdot \log \left(a\right)$ and transform the left expression:
$2 \log \left(x\right) = \log \left({x}^{2}\right)$

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, $\log \left(f \left(x\right)\right) < \log \left(g \left(x\right)\right) \iff f \left(x\right) < g \left(x\right)$ holds.

$\log \left({x}^{2}\right) < \log \left(2 x - 1\right)$
$\iff {x}^{2} < 2 x - 1$
$\iff {x}^{2} - 2 x + 1 < 0$
$\iff {\left(x - 1\right)}^{2} < 0$
Now, this is a false statement for all real numbers $x$ since a quadratic expression is always $\ge 0$.
This means that (under the assumption that your logarithm basis is indeed $> 1$) your inequality has no solutions.