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avatar+211 

Solve the system of equations
y = \log_2 (2x)
y = log_4 (16 + x)

 May 24, 2024
 #1
avatar+806 
0

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! :)

 May 24, 2024
 #2
avatar+129495 
+1

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

 

cool cool cool

 May 24, 2024

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