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$log(2x-1)+2log\sqrt{x-9}\geq2$

Using properties of logs we can write it as

$log(2x-1)+log({x-9})\geq2$

$\rightarrow log(2x-1)(x-9)\geq log100$

$\rightarrow (2x-1)(x-9)\geq100$

$\rightarrow 2x^{2}-19x-91\geq0$

$\rightarrow(2x+7 )(x-13)\geq0$

We can also see that $x$ can't be less than or equal to $\dfrac {1}{2}$ because log for -ve no.s is not defined.

By this $x\in [13 \ , \ \infty )$

$\huge{\rightarrow |x|_{min}=13}$

The question is flawed as it does not specify the base to which the logarithm is specified. A more general solution would proceed thus :

$\frac{log(2x-1)}{2}+log(\sqrt{x-9}) \geq 1$

From this, it is clear that $x>9$ for the square root to exist. Which translates to

$log(2x-1)+log(x-9) \geq 2$

If it is assumed that the base is an unspecified positive number $k$ , then the above inequality is equivalent to

$(2x-1)(x-9) \geq k^2$

or

$2x^2-19x+9-k^2 \geq 0$

Which gives the solution as

$x \leq \frac{1}{4}(19-\sqrt{8k^2+289}$ or $x \geq \frac{1}{4}(19+\sqrt{8k^2+289}$

The first inequality is valid only if $19>\sqrt{8k^2+289}$ or |(x<3). However, $x>9$ . Hence, the second inequality would yield the solution as

$x \ge ceil\left(\frac{1}{4}(19+\sqrt{8k^2+289}\right)$ where $ceil(y)$ is used to denote the least integer greater than $y$

If natural logarithm is assumed then

$\huge{x\geq10}$ .

However, if the base is assumed to be 10, the solution would be

$\huge{x \geq 13}$ .

If the base is assumed to be 100 (just for argument sake), then the solution would be

$\huge{x \geq 76}$