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.

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
$a^3=\frac{4}{3}$.
