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

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Answer 1 · 2017-10-16T09:17:38

$x approx +- 1.822616$

$(2^x)^x = 10$

$2^(x^2) =10$

$x^2 log_10 2= 1$

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

$x approx +- sqrt 3.321928$

$approx +- 1.822616$

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