Arithmetic progression of logarithms 
If $\log_{\sqrt{2}}a$, $\log_{\sqrt{2}}(2a^2)$, $\log_{\sqrt{2}}(a^3+4)$ are in A.P, find the value of "$a$".

I tried solving this and I am getting $a^3 = \frac{4}{3}$ is it correct. Please guide.
 A: $$\log_{\sqrt{2}}a-\log_{\sqrt{2}}(2a^2)=\log_{\sqrt{2}}(2a^2)-\log_{\sqrt{2}}(a^3+4)$$
$$a/(2a^2)=2a^2/(a^3+4)$$ 
$$a(a^3+4)=4a^4$$
$$3a^4-4a=0$$
$$a(3a^3-4)=0$$
0nly$$a^3=4/3$$ is solution
A: If $$\log_{\sqrt{2}}a, \log_{\sqrt{2}}(2a^2), \log_{\sqrt{2}}(a^3+4)$$ are in A.P,
$$\log_{\sqrt{2}}(2a^2)-\log_{\sqrt{2}}a=\log_{\sqrt{2}}(a^3+4)-\log_{\sqrt{2}}(2a^2)$$
$$\implies \log_{\sqrt{2}}a+\log_{\sqrt{2}}(a^3+4)=2\log_{\sqrt{2}}(2a^2)$$
$$\implies \log_{\sqrt{2}}a(a^3+4)=\log_{\sqrt{2}}(2a^2)^2$$
$$\implies a(a^3+4)=(2a^2)^2\implies 3a^4=4a\implies 3a^3=4\text { as } a\ne0$$
A: You are correct. Because the logarithms are in arithmetic progressions, there is some $d$ such that
$$\log_{\sqrt{2}}(2a^2)=\log_{\sqrt{2}}(a)+d$$
and
$$\log_{\sqrt{2}}(a^3+4)=\log_{\sqrt{2}}(a)+2d.$$
We know that 
$$\log_{\sqrt{2}}(2a^2)=\log_{\sqrt{2}}(2)+2\log_{\sqrt{2}}(a)=2+2\log_{\sqrt{2}}(a)$$
so $\log_{\sqrt{2}}(a)=d-2$, and hence $\log_{\sqrt{2}}(2a^2)=2d-2$. Therefore $$\log_{\sqrt{2}}(2a)=\log_{\sqrt{2}}(2a^2)-\log_{\sqrt{2}}(a)=d,$$
hence
$$\log_{\sqrt{2}}(a^3+4)=\log_{\sqrt{2}}(a)+2\log_{\sqrt{2}}(2a)=\log_{\sqrt{2}}(4a^3),$$
hence
$$a^3+4=4a^3,$$
and therefore
$a^3=\frac{4}{3}$.
A: So the difference between the first two equals the difference between the next two:
$$
\log_{\sqrt{2}}(2a^2)-\log_{\sqrt{2}} a = \log_{\sqrt{2}}(a^3+4)-\log_{\sqrt{2}}(2a^2).
$$
$$
\log_{\sqrt{2}}(a^3+4) - 2\log_{\sqrt{2}}(2a^2) + \log_{\sqrt{2}} a = 0.
$$
$$
\log_{\sqrt{2}} \left( \frac{a(a^3+4)}{(2a^2)^2} \right) = 0.
$$
$$
\frac{a(a^3+4)}{(2a^2)^2} = 1.
$$
$$
a^3+4 = 4a^3.
$$
$$
4 = 3a^3.
$$
One way of looking at the problem is to say that whatever you add to $\log_{\sqrt{2}} a$ to get $\log_{\sqrt{2}} (2a^2)$ is the same thing that you add to $\log_{\sqrt{2}} (2a^2)$ to get $\log_{\sqrt{2}}(a^3+4)$, and that implies that whatever you multiply $a$ by to get $2a^2$, that's the same thing you multiply $2a^2$ by to get $a^3+4$.  But the thing you multiply $a$ by to get $2a^2$ is $2a$.  That means multiplying $2a^2$ by $2a$ should get you $a^3+4$.  Hence $4a^3=a^3+4$.
