Help meee

First off, we can subsitue y out to get \(\log _4\left(16+x\right)=\log _2\left(2x\right)\). 

Now, we just have to solve for x. \(\log _4\left(16+x\right)=\log _2\left(x\right)+1\). 

I'm not sure exactly how to explain, but now we just have to isolate x and we eventually get

\(x=\frac{1+\sqrt{257}}{8}\)

Here's a graph to make it more clear.

Their intersection point is our answer for x! 

Thanks! :)

log₂ (2x)  = log₄ (16 + x)

Using the change-of-base theorem

log (2x) / log2  = log (16 + x) / log 4

log (2x) / log 2 = log (16 + x) / log 2^2

log (2x) / log 2  = log (16 + x) / [ 2 log 2]      multiply through by log 2

log (2x)  = log (16 + x) / 2

2log (2x) = log (16 + x)

log (2x)^2 = log (16 + x)      this implies that

(2x)^2 = 16 + x

4x^2 - x - 16  =  0

Using the Quadratic Formula  (and taking the positive value of x )

x =  [ 1 + sqrt [ 1^2 - 4 (4) (-16) ] ] / (2 *4)

x = [ 1 + sqrt (257) ] / 8