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

 Mar 28, 2024
 #1
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y = log 2 (2x)

y= log 4 (16 + x)

 

Change of base theorem

 

log ( 2x) / log 2  =  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 both sides by log 2 }

 

log (2x)  =  log (16 + x) / 2

 

2 log (2x)  = log (16 + x)

 

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

 

2x^2  = 16 + x

 

4x^2  - x  - 16  =  0        { x must be positive }

 

x = [ 1 + sqrt [ 1 + 256 ] ]  / 8  =   [ 1 + sqrt (257) ] /  8 

 

cool cool cool

 Mar 28, 2024

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