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**

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