For which $x$ is
$$\log(2),\log(2^x-1),\log(2^x+3),\ldots$$
an arithmetic sequence?
Solved: For $d=\log(2)$, one gets the airthmetic sequence $$n\log(2)$$
Then you have to solve $$4=2^x-1$$
This means $2^x=5$, from which one attains $$8=2^x+3$$
From $2^x=5$ and $\log(a^b)=b\log(a)$, deduce that
$$x=\frac{\log(5)}{\log(2)}$$
$\log(2)+\log(2^x+3)=2\log(2^x-1)=\log(2^x-1)^2.$
$2(2^x+3)=(2^x-1)^2= 2^{2x}-2\times2^x+1.$
$2^{2x}-4\times2^x-5=0.$
$(2^{x}-5)\times(2^x+1)=0.$
$2^{x}=5.$
$2^x=-1.$