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Find the positive value of x that satisfies \(4^{\log_2 x} + x^2 = 8\)

 Dec 11, 2019
 #1
avatar+33603 
+3

From inspection it's clear that x = 2 satisfies the equation.

 Dec 11, 2019
 #2
avatar+128053 
+1

4 log 2 x   +   x^2  =   8           we can write

 

(2^2)^(log2 x)  + x^2  =  8

 

2^(2 log 2x)  +  x^2   = 8

 

2^( log 2 x^2)  + x^2  =  8

 

[ Note that we have the property that    a ^ (log a  b^2)  =  b^2   ]

 

x^2   + x^2   =   8

 

2x^2  =  8                  divide both sides by 2

 

x^2  =  4                take the positive root

 

x   = 2

 

 

cool cool cool

 Dec 11, 2019

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