Question

Solve the following equation:

`3^(2x + 3) - 28.3^x + 1 = 0`

Answer 1

Given the equation: `3^(2x + 3) - 28.3^x + 1 = 0`

Step-wise calculation:

1. Note that

`3^(2x + 3) = 3^3 xx 3^(2x)`

`3^(2x + 3) = 27 xx (3^x)^2`

2. Let (y = 3^x), then the equation becomes:

27y² – 28y + 1 = 0

3. This is a quadratic in (y):

27y² – 28y + 1 = 0

4. Solve this quadratic equation using the quadratic formula:

`y = (28 +- sqrt((-28)^2 - 4 xx 27 xx 1))/(2 xx 27)`

`y = (28 +- sqrt(784 - 108))/54`

`y = (28 +- sqrt(676))/54`

`y = (28 +- 26)/54`

5. Two possible values for (y):

`y_1 = (28 + 26)/54`

`y_1 = 54/54`

y₁ = 1

`y_2 = (28 - 26)/54`

`y_2 = 2/54`

`y_2 = 1/27`

6. Recall (y = 3^x), so:

If 3^x = 1, then x = 0

If `3^x = 1/27 = 3^(-3)`, then x = –3

The solutions to the equation are x = 0 or x = –3.