Given the logarithm equation:

ln x^3 = 2 + ln x

We need to fin x value that satisfies the equation.

First we will subtract ln x from both sides.

==> ln x^3 - ln x = 2

Now we will use the logarithm properties to solve.

We know that ln a - ln b = ln a/b

==> ln x^3 - ln x = ln x^3/x = ln x^2 = 2

==> ln x^2 = 2

Now we will rewrite using the exponent form.

==> x^2 = e^2

Since we have the powers are equal, then the bases are equal too.

**==> x = e **

We have to solve for x given that ln x^3 = 2 + ln x

using the exponential rule for logarithms

ln x^3 = 2 + ln x

=> 3* ln x =2 + ln x

=> 3*ln x - ln x = 2

=> 2* ln x = 2

=> ln x = 1

As the base of ln is e x = e.

**Therefore x = e.**

lnx^3 = 2+lnx. To solve for x.

=> 3lnx = 2+lnx, lnx^3 = 3lnx, as lna^m = m lnx.

=> 3lnx- lnx = 2.

=> 2lnx = 2

=> (2lnx)/2 = 2/2 = 1.

=> lnx = 1.

=> x = e = 2.718281828.

lnx^3 = 2+lnx. To solve for x.

=> 3lnx = 2+lnx, lnx^3 = 3lnx, as lna^m = m lnx.

=> 3lnx- lnx = 2.

=> 2lnx = 2

=> (2lnx)/2 = 2/2 = 1.

=> lnx = 1.

=> x = e = 2.718281828.

lnx^3 = 2+lnx. To solve for x.

=> 3lnx = 2+lnx, lnx^3 = 3lnx, as lna^m = m lnx.

=> 3lnx- lnx = 2.

=> 2lnx = 2

=> (2lnx)/2 = 2/2 = 1.

=> lnx = 1.

=> x = e = 2.718281828.