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How do I solve #(2^x)^x=10#?

1 Answer

Oct 16, 2017

#x approx +- 1.822616#

Explanation:

#(2^x)^x = 10#

#2^(x^2) =10#

#x^2 log_10 2= 1#

#x^2 = 1/log_10 2 approx 3.321928#

#x approx +- sqrt 3.321928#

# approx +- 1.822616#

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