How do you solve #2lnx+3ln2=5#?
1 Answer
Dec 15, 2015
Explanation:
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^(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.
