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Solve the system of equations

y = log_2 (2x)

y = log_4 (16 + x)

 Dec 29, 2023
 #1
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y = log 2 (2x)

y = log 4 (16 + x)

 

log 2 (2x)  =  log 4 (16 + x)         {change of base)

 

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]

 

log (2x)  = log (16 + x) / 2

 

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

 

log (2x)^2  = log (16 + x)

 

4x^2  = 16 + x

 

4x^2 - x - 16  =  0

 

x must be positive.....so......

 

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

 

 

cool cool cool

 Dec 30, 2023

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