# 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.