If $\log 2, \log (2x - 1), \log (2x + 3)$ are in A.P, then $x$ is equal to ____

When $a,b,c$ are in A.P, $2b=a+c$

Hence,

$2[log(2x-1)]=log(2)+log(2x+3)$

We know that $n.log(y)=log(y)^n$ and $log(a)+log(b)=log(ab)$

So, $log(2x-1)^2=log(4x+6)$

So, $(2x-1)^2=4x+6$

$4x^2+1-4x=4x+6$

$4x^2-8x-5=0$

$4x^2-10x+2x-5=0$

$2x(2x-5)+1(2x-5)=0$

$(2x+1)(2x-5)=0$

$x=-\frac{1}{2}$ or $\frac{5}{2}$

But log values are always positive, hence $x=\frac{5}{2}$