(1/9)^x-2(1/3)^x-1-27=0

3 min read Jun 07, 2024

Solving the Equation: (1/9)^x - 2(1/3)^x - 1 - 27 = 0

In this article, we will explore how to solve the equation (1/9)^x - 2(1/3)^x - 1 - 27 = 0. This equation involves exponential functions and requires some clever manipulation to solve.

Step 1: Simplify the Equation

First, let's simplify the equation by noticing that (1/9) = (1/3)^2. So, we can rewrite the equation as:

((1/3)^2)^x - 2(1/3)^x - 1 - 27 = 0

Using the rule of exponents, we can rewrite the equation as:

(1/3)^(2x) - 2(1/3)^x - 1 - 27 = 0

Step 2: Substitute y = (1/3)^x

Now, let's substitute y = (1/3)^x into the equation. This gives us:

y^2 - 2y - 28 = 0

Step 3: Factor the Quadratic Equation

Now, we have a quadratic equation in the form of ax^2 + bx + c = 0. We can factor the equation as:

(y + 4)(y - 7) = 0

Step 4: Solve for y

From the factored equation, we can see that either (y + 4) = 0 or (y - 7) = 0.

Solving for the first factor, we get:

y + 4 = 0 --> y = -4

And solving for the second factor, we get:

y - 7 = 0 --> y = 7

Step 5: Solve for x

Now, we can substitute y = (1/3)^x back into the solutions we found for y. This gives us:

(1/3)^x = -4

and

(1/3)^x = 7

Taking the logarithm of both sides of each equation, we get:

x log(1/3) = log(-4)

and

x log(1/3) = log(7)

Dividing both sides by log(1/3), we get:

x = log(-4) / log(1/3)

and

x = log(7) / log(1/3)

Therefore, we have found the solutions to the equation (1/9)^x - 2(1/3)^x - 1 - 27 = 0.