# 3 ln x - ln 3 = 3

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3ln x - ln 3 = 3

We know that a ln b = ln b^a

==> ln x^3 - ln 3 = 3

Also, we know that ln a - ln b = ln (a/b)

==> ln (x^3/3) = 3

==> x^3/3 = e^3

==> x^3 = 3e^3

**==> x = (3)^1/3 * e**

First, we'll impose the constraint of existence of logarithms:

x>0

Now, we'll use the power property of logarithms, so that:

3 ln x = ln x^3

We'll re-write the equation:

ln x^3 - ln 3 = 3

Since the bases of logarithms are matching, we'll use the quotient rule:

ln (x^3/3) = 3

We'll have:

x^3/3 = e^3

x^3 = 3e^3

**x = e*[(3)^1/3**]

**Since the solution is positive, it is valid.**

3lnx -ln3 = 3

Solution;

3lnx - ln3 =lne^3 by definition

3lnx = ln3+lne^3.

3lnx = ln3*e^3. as lna+lnb = lnab

lnx^3 = ln3*e^3, as m ln a = ln a^m

Take antilog of both sides:

x^3 = 3e^3

x = (3e^3)^(1/3)

x = {(3^(1/3)}e

lnx^3-ln3=3 ln(x^3/3)=3

x^3/3=e^3 x^3=3e^3 x=e*3^1/3