How do you solve #7+ 27\ln x = 4#?

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Answer 1 · 2017-12-10T10:39:43

# x = e^(-1/9) #

The first thing is to rearange:

#27lnx = -3 #

Dividing through by 27:

#=> lnx = -1/9 #

Raising each side by #e#:

#=> e^(lnx) = e^(-1/9) #

Using our logarithms knowledge: # alpha^(log_alpha beta) = beta #

Hence:

#=> x = e^(-1/9) approx 0.90#