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Solve the following exponential equations:

 

c. (4)^x-3  = 1/16

 

d. 3^x^2-42 = 1/3^x

 

f. x^2 2^x - 4(2^x) = 0

 Feb 23, 2019
 #1
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(c)

\(4^{x-3} = \dfrac{1}{16}\\ 4^{x-3} = 4^{-2}\\ x -3=-2\\ x=1\)

 

(d)

\(3^{x^2-42} = \left(\dfrac{1}3\right)^x\\ 3^{x^2-42} = 3^{-x}\\ x^2-42=-x\\ x^2+x-42=0\\ (x+7)(x-6)=0\\ x = -7\text{ or }x=6\)

 

(f)

\(x^22^x-4(2^x)=0\\ 2^x(x^2-4)=0\\ 2^x(x-2)(x+2) =0\\ 2^x =0\text{(rejected) or } x = \pm 2\\ x = \pm 2\)

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 Feb 23, 2019

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