Please help to solve....?

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2 Answers
Oct 27, 2017

# 25 sqrt5#

Explanation:

#4^x-1/4 4^x = 3 * 8# or

#4^x(1-1/4)=4^x(3/4)=3*8rArr 4^x = 4*8 = 2^5# then

#2^(2x) = 2^5 rArr 2x=5 rArr x = 5/2#

and finally

#(2x)^x = (2*5/2)^(5/2) = sqrt(5^5)=5^2sqrt5 = 25 sqrt5#

Oct 27, 2017

(c) #25sqrt5#

Explanation:

#4^x - 4^(x-1) = 24#

#4^x(1-1/4) =24#

#4^x(3/4) =24#

#4^x = (24xx4)/3 =32#

#2^(2x) = 2^5#

Equating exponents:

#2x=5 -> x=2.5#

We are asked to evaluate #(2x)^x#

#(2x)^x = (2xx2.5)^(5/2)#

#= 5^(5/2) = sqrt(5xx5xx5xx5xx5)#

#= 5xx5xxsqrt5#

#= 25sqrt5# which is answer (c)