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Find the sum of all values of x such that $2^{x^2-3x-2} = 4^{x - 4} \cdot 8^x$.

 Jul 27, 2021
 #1
avatar+2193 
+1

2^(x^2 - 3x - 2) = 2^(2*(x-4)) * 2^(3x)

2^(x^2 - 3x - 2) = 2^(5x - 8)

x^2 - 3x - 2 = 5x - 8

x^2 - 8x + 6 = 0

Using vieta's, -b/a = --8/1 = 8.

 

=^._.^=

 Jul 27, 2021
 #2
avatar+34315 
+1

2^(x2-3x-2)  = (2^2)x-4  * 2^3x     equate the exponents

 

x^2 -3x -2 = 2x-8+3x

x^2 -8x +6     Use quadratic formula to find   x =   4 +- sqrt(10)      summed = 8

 Jul 27, 2021
edited by ElectricPavlov  Jul 27, 2021

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