How do you solve #2lnx+3ln2=5#?

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Answer 1 · 2015-12-15T08:00:19

#x ~= 4.30716#

Property of Logarithmic expression

#log A + log B = Log(AB) " " " " " (1)#
#n log A= log A^n " " " " (2)#

Given :

#2ln x + 3ln 2 = 5#

Rewrite as:
Using rule (1)

#lnx^2 + ln2^3 = 5#

Using rule (1)

#ln(x^2 * 8) = 5#

Raise the expression to exponential form, with the base of #e#

#e^(ln(8x^2) = e^5#

#8x^2 = e^5#
#x^2 = (e^5)/8#

#x = +-sqrt((e^5)/8)#

#x ~= 4.30716#

Because the argument of any logarithm always POSITIVE and greater than zero, due to domain restriction.