How do you solve for x in #7.316 = e^(ln(2x))#?

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Answer 1 · 2016-07-06T18:10:06

# x = 3.658#

#ln(x) = log_e(x)#

The function #ln(x) = a# finds the value of a such that

#x = e^a#

Therefore, #e^(ln(x)) = e^a = x#

What I'm trying to say is that #e^x# and #ln(x)# are inverse operations, so they cancel out.

#e^(ln(2x)) = 2x#

So #7.316 = 2x implies x = 3.658#