Given the equation e^(ln 2 + 3 ln x)

We need to simplify.

Let y= e^(ln 2 + 3lnx)

We know that a*lnb = ln b^a

Then we will rewrite 3ln x = ln x^3

==> y= e^(ln 2 + ln x^3)

Now, we know that ln a + ln b = ln a*b

Then, we will rewrite ln 2 + ln x^3 = ln 2x^3

==> y= e^(ln 2x^3)

Now, we know that e^ln x = x

==> y= 2x^3

Then we conclude that:

**e^(ln 2 + 3lnx) = 2x^3**

We have to simply e^ (ln 2 + 3 ln x)

Here we use the properties that d^ (a + b) = d^a * d^b

e^ (ln 2 + 3 ln x)

=> e^ (ln 2) * e^ (3 ln x)

a log b = log a^b

=> e^ (ln 2) * e^ (ln x^3)

Now for any logarithm to a base a x^ (log (x) b) = b

=> 2 * x^3

**Therefore e^ (ln 2 + 3ln x) = 2*x^3**

We'll use the property of multiplication of 2 exponentials, having matching bases.

e^x*e^y = e^(x+y)

We'll compare e^(x+y) and e^( ln 2 + 3ln x) and we'll get:

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

But 3 ln x = ln x^3 (the power rule of logarithms)

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

But e^ ln a = a

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

So, the simplified expression is:

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

To simplify e^( ln 2 + 3ln x).

e^(ln2+3lnx) = e^(ln2+lnx^3), as mlna= lna^m.

e^(ln2+3lnx) = e^ln(2*x^3), as lna +lnb = ln (a*b).

e^(ln2+3lnx) = 2x^3, as e^lna = y implies a = y.

Therefore e^(ln2+3lnx) = 2x^3.