<s>[INST] 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.
 [/INST] 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 $4a^3=a^3+4$.
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