# ln 3x + ln e^2 = 2

hala718 | Certified Educator

ln 3x + ln e^2 = 2

We know that:

ln e^a = aln e = a

==> ln 3x + 2 = 2

Now subtract 2 from both side:

==> ln 3x = 0

==> 3x = e^0 = 1

==> x = 1/3

giorgiana1976 | Student

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.

neela | Student

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.