# Hi there, can anyone help me with the following equation. 2e^3x - 15e^-3x = -1 i know I have to convert it to a quadratic, but not sure how. thanks

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### 1 Answer

You need to use the following rule, such that:

`e^(-f(x)) = 1/(e^(f(x)))`

Replacing 3x for f(x) yields:

`e^(-3x) = 1/(e^(3x))`

Hence, you can write the equation, such that:

`2e^(3x) - 15/(e^(3x)) = -1`

You need to make the following substitution, such that:

`e^(3x) = y`

Replacing y for e`^(3x)` , in equation, yields:

`2y - 15/y = -1`

You need to multiply each term by y, such that:

`2y^2 - 15 = -y`

You need to move all terms to the left, such that:

`2y^2 + y - 15 = 0`

You should use quadratic formula, such that:

`y_(1,2) = (-1+-sqrt(1^2 - 4*2*(-15)))/(2*2)`

`y_(1,2) = (-1+-sqrt121)/4`

`y_1 = (-1+11)/4 = 10/4 = 5/2`

`y_2 = (-1-11)/4 = -3`

You may evaluate `x_(1,2)` , such that:

`e^(3x_1) = y_1 => e^(3x_1) = 5/2`

You need to take logarithms both sides, such that:

`ln e^(3x_1) = ln 5/2`

`3x_1*ln e = ln 5/2`

`3x_1*1 = ln 5/2`

`x_1 = (1/3)* ln 5/2 => x_1 = ln (5/2)^(1/3) => x_1 = ln root(3)(5/2)`

You need to find `x_2` , such that:

`e^(3x_2) = y_2 => e^(3x_2) =-3`

The equation `e^(3x_2) =-3` is not valid, since` e^(3x_2) > 0` , hence `e^(3x_2) ` cannot be negative 3.

**Hence, evaluating the solution to the given equation, using quadratic equation, yields **`x_1 = ln root(3)(5/2).`