Logarithms in AP

$\begin{aligned} S & = \log_2 4 + \log_2 16 + \log_2 64 + \cdots + x \\ & = \log_2 2^2 + \log_2 2^4 + \log_2 2^6 + \cdots + \color{#3D99F6}x & \small \color{#3D99F6} \text{Let }x = 2n \\ & = 2 + 4 + 6 + \cdots + 2n \\ & = 2(1+2+3+\cdots + n) \\ & = n(n+1) \end{aligned}$

Since $S = 42$ :

$\begin{aligned} n(n+1) & = 42 = 6(6+1) \\ \implies n & = 6 \\ x & = 2n = 12 & \small \color{#3D99F6} \text{Since }x = \log_2 a \\ \log_2 a & = 12 \\ \implies a & = 2^{12} = \boxed{4096} \end{aligned}$

The first three logs are 2,4,6,... which are just the whole numbers doubled. 42 is 21 doubled and we know that the sixth triangular number is 1+2+3+4+5+6=21. So we need the last log to be the sixth even number: 12.

$2^{12}=\boxed{4096}$

2 + 4 + 6 + 8 + 10 + 12 = 42. X is hence log 2 to the base A. Hence a = 2 to the power 12 = 4096