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Nov 20, 2018 · The first equation gives x=2 x = 2 , the second one gives x=1 x = 1 . Therefore, the set of possible solutions is S={1,2} S = { 1 , 2 } .
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X=5 PREMISES 2^x-32=0 CALCULATIONS 2^x-32=0 Converting bases, 2^x-2^5=0 2^x=2^5 Evaluating, x=5 C.H.. Upvote ·. Comments. About the Author.
Aug 17, 2021 · If I understand you, you want the value of x in 2^x=32. Then x = 5 since 2^5 means 2*2*2*2*2=32.
Feb 29, 2024 · Define y = 2^x. Then, you have. (1/8)y^2 + 2y + 32 = 0. y^2 + 16y + 256 = 0. There are no real solutions for y or x. 162 views · ·.
Feb 29, 2024 · Define y = 2^x. Then, you have. (1/8)y^2 + 2y + 32 = 0. y^2 + 16y + 256 = 0. There are no real solutions for y or x. 115 views · ·.
Jul 27, 2020 · Therefore, the solution to the equation is x = -1. Therefore, the value of x in the expression (2x^2+4x+2=0) is -1.
Feb 19, 2021 · You might try by calling 2^x = t , then 4^x = 2^(2x ) =( 2^x)^2 = t^2 and you will have a quadratic in t: 32 t^2 - 97t + 3 = 0 which can be ...