Before solving the equation, we'll have to impose constraints of existence of logarithms.

3x>0

We'll divide by 3:

x>0

Now, we'll apply the product property of logarithms:

ln 3x + ln e^2 = ln 3x*e^2

ln 3x*e^2 = 2

We could write, using the power property of logarithms, that:

2 = 2*1 = 2*ln e = ln e^2

ln 3x*e^2 = ln e^2

Because the logarithms from both sides have matching bases, we'll use the one to one property:

3x*e^2= e^2

We'll divide by e^2 both sides:

**3x = 1**

**x = 1/3>0**

Because the solution is in the interval (0, +inf.), the solution is admissible.

To solve ln3x+lne^2=2

Solution

By property of logarithms log ab = loga+logb. ln e^m = m* ln e = m* 1 =m

So ln3x = ln3 +lnx . lne^2 = 2*1 =2.

So we rewrite the LHS of the given equation as:

ln3+ lnx + 2 = 2

lnx = -ln3 = (-1)*ln3 = ln (3^-1) = ln(1/3)

lnx = ln (1/3). Take antilogarithms.

x =1/3.