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.

If $$\log_{\sqrt{2}}a, \log_{\sqrt{2}}(2a^2), \log_{\sqrt{2}}(a^3+4)$$ are in A.P,
$$\log_{\sqrt{2}}(2a^2)-\log_{\sqrt{2}}a=\log_{\sqrt{2}}(a^3+4)-\log_{\sqrt{2}}(2a^2)$$
$$\implies \log_{\sqrt{2}}a+\log_{\sqrt{2}}(a^3+4)=2\log_{\sqrt{2}}(2a^2)$$
$$\implies \log_{\sqrt{2}}a(a^3+4)=\log_{\sqrt{2}}(2a^2)^2$$
$$\implies a(a^3+4)=(2a^2)^2\implies 3a^4=4a\implies 3a^3=4\text { as } a\ne0$$
