We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+1
176
1
avatar+123 

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
avatar+7725 
+1

(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)

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

.
 Feb 23, 2019

37 Online Users