# How do you solve for x the exponential?e^3x*e^2x-3=2

### 2 Answers | Add Yours

We have to determine x given that e^3x*e^2x - 3 = 2

e^3x*e^2x - 3 = 2

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

=> e^3x*e^2x = 5

use the relation x^a*x^b = x^(a + b)

=> e^(3x + 2x) = 5

=> e^5x = 5

Take the natural log of both the sides

ln [ e^5x] = ln 5

=> 5x = ln 5

=> x = (ln 5)/5

**The value of x = (ln 5)/5**

By definition, the superscripts of two exponentials that have matching bases and they are multiplied, must be added.

Therefore, (e^3x)*(e^2x) = e^(3x+2x) = e^(5x)

The equation will become:

e^(5x) - 3 = 2

e^(5x) = 3 + 2

e^(5x) = 5

We'll take natural logarithms both sides:

ln e^(5x) = ln 5

We'll apply the power rule:

5x*ln e = ln 5

But ln e = 1

5x = ln 5

x = ln 5/5

**The solution of the equation is x = (ln 5)/5.**