Question #0392a

1 Answer
Cesareo R.
Dec 20, 2017

See below.

Explanation:

Assuming #log x = log_e x#

#3 = e^lambda# with #lambda = log_e 3# and then

#e^(lambda log_e x) = x^lambda# then

#x^lambda +3 x^lambda = 2 rArr x^lambda = 1/2# or

#x = (1/2)^(1/lambda)= (1/2)^(1/log_e 3) approx 0.532097#

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