How do we solve #sqrt(2^x)=4096#?

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Answer 1 · 2017-03-24T04:32:15

#x=24#

#sqrt(2^x)=4096#, but as #4096=2^12#

#(2^x)^(1/2)=2^12#

i.e #2^((x xx1/2))=2^12#

or #2^(x/2)=2^12#

i.e. #x/2=12#

and #x=12xx2=24#

Answer 2 · 2017-03-24T11:33:25

#x = 24#

It will be to your advantage to learn some of the common powers by heart. The powers of 2 are well worth knowing.

#4096 = 2^12#

#sqrt(2^x) = 4096#

#sqrt(2^x) = 2^12" "larr# square both sides to get rid of the root

#(sqrt(2^x))^2 = (2^12)^2#

#2^x = 2^12" "larr# the bases are equal, so the indices are equal

#:. x = 24#

Answer 3 · 2017-03-24T11:51:17

#x = 24#

We have: #(sqrt(2))^(x) = 4096#

Let's express all numbers in terms of #2#:

#Rightarrow (2^((1) / (2)))^(x) = 2^(12)#

Using the laws of exponents:

#Rightarrow 2^((1) / (2) x) = 2^(12)#

#Rightarrow (1) / (2) x = 12#

Finally, to solve for #x#, let's divide both sides of the equation by #(1) / (2)#:

#Rightarrow ((1) / (2) x) / ((1) / (2)) = 12 / ((1) / (2))#

#therefore x = 24#