Solving $(2^{2x+1})(3^{2x+2})+2^x(3^{x+2})-2=0$ Exponential Equation

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Introduction

In this article, we will delve into the intricacies of solving the exponential equation $(2^{2x+1})(3^{2x+2}) + 2^x(3^{x+2}) - 2 = 0$. Exponential equations are a fascinating area of mathematics, often requiring clever algebraic manipulations and a solid understanding of exponent rules to solve. This particular equation presents a unique challenge due to the combination of different bases (2 and 3) and the presence of multiple terms. Our approach will involve simplifying the equation, making a suitable substitution to transform it into a more manageable form, and then solving for the unknown variable, x. We will explore each step in detail, providing explanations and justifications to ensure a comprehensive understanding of the solution process. The goal is not just to find the answer, but to equip you with the skills and knowledge to tackle similar exponential equations in the future. Let's begin by dissecting the equation and identifying potential strategies for simplification.

Simplifying the Equation

To begin, let's rewrite the given equation:

$(2^{2x+1})(3^{2x+2}) + 2^x(3^{x+2}) - 2 = 0$

Our first step in simplifying this equation involves using the properties of exponents to break down the terms. Recall that $a^{m+n} = a^m imes a^n$. Applying this rule, we can rewrite the terms as follows:

$2^{2x+1} = 2^{2x} imes 2^1 = 2 imes 2^{2x} = 2 imes (2^2)^x = 2 imes 4^x$

$3^{2x+2} = 3^{2x} imes 3^2 = 9 imes 3^{2x} = 9 imes (3^2)^x = 9 imes 9^x$

$3^{x+2} = 3^x imes 3^2 = 9 imes 3^x$

Substituting these simplified expressions back into the original equation, we get:

$(2 imes 4^x)(9 imes 9^x) + 2^x(9 imes 3^x) - 2 = 0$

Now, let's simplify further by multiplying the constants:

$18 imes 4^x imes 9^x + 9 imes 2^x imes 3^x - 2 = 0$

Notice that $4^x imes 9^x$ can be written as $(4 imes 9)^x = 36^x$, and $2^x imes 3^x$ can be written as $(2 imes 3)^x = 6^x$. Substituting these back into the equation, we obtain:

$18 imes 36^x + 9 imes 6^x - 2 = 0$

Now we have a more simplified form of the equation. The next step is to recognize a potential substitution that will help us transform this exponential equation into a more manageable algebraic equation. This involves identifying a common base or a relationship between the exponential terms. In this case, we can see that $36^x$ is related to $6^x$ since $36 = 6^2$. This observation will guide our next step in solving the equation.

Making a Substitution

Having simplified the equation to $18 imes 36^x + 9 imes 6^x - 2 = 0$, the next logical step is to make a substitution to further simplify it. We observe that $36^x$ can be expressed as $(6^2)^x$, which is equivalent to $(6^x)^2$. This suggests that we can use $6^x$ as our substitution variable.

Let's introduce a new variable, say y, such that:

$y = 6^x$

Then, $36^x$ can be rewritten as:

$36^x = (6^2)^x = (6^x)^2 = y^2$

Now, we can substitute y and $y^2$ into our equation:

$18y^2 + 9y - 2 = 0$

This substitution has transformed our exponential equation into a quadratic equation in terms of y. Quadratic equations are a familiar territory, and we have various methods at our disposal to solve them, such as factoring, completing the square, or using the quadratic formula. This substitution is a crucial step in solving the original exponential equation, as it allows us to apply algebraic techniques that are not directly applicable to exponential forms. The resulting quadratic equation is much easier to handle, and we can now proceed to solve for y.

Solving the Quadratic Equation

We have now transformed the original exponential equation into a quadratic equation: $18y^2 + 9y - 2 = 0$. To solve for y, we can use a variety of methods, such as factoring, completing the square, or applying the quadratic formula. In this case, let's try factoring the quadratic expression.

We are looking for two numbers that multiply to $(18)(-2) = -36$ and add up to 9. These numbers are 12 and -3. So, we can rewrite the middle term (9y) as 12y - 3y:

$18y^2 + 12y - 3y - 2 = 0$

Now, we can factor by grouping:

$6y(3y + 2) - 1(3y + 2) = 0$

Notice that we have a common factor of $(3y + 2)$. Factoring this out, we get:

$(6y - 1)(3y + 2) = 0$

For the product of two factors to be zero, at least one of the factors must be zero. Therefore, we have two possible cases:

1. $6y - 1 = 0$
2. $3y + 2 = 0$

Solving the first case for y, we get:

$6y = 1$

y = rac{1}{6}

Solving the second case for y, we get:

$3y = -2$

y = -rac{2}{3}

So, we have found two possible values for y: rac{1}{6} and -rac{2}{3}. However, we must remember that y was a substitution variable, and we need to relate these values back to the original variable x. In the next step, we will substitute these values back into the equation $y = 6^x$ and solve for x.

Substituting Back and Solving for x

Having found the values of y from the quadratic equation, which are y = rac{1}{6} and y = -rac{2}{3}, we now need to substitute these back into the original substitution equation $y = 6^x$ to solve for x.

Let's consider the first value, y = rac{1}{6}:

rac{1}{6} = 6^x

We can rewrite rac{1}{6} as $6^{-1}$, so the equation becomes:

$6^{-1} = 6^x$

Since the bases are the same, we can equate the exponents:

$x = -1$

Now, let's consider the second value, y = -rac{2}{3}:

-rac{2}{3} = 6^x

Here, we encounter a problem. The exponential function $6^x$ is always positive for any real value of x. It is impossible for $6^x$ to be equal to a negative number such as -rac{2}{3}. Therefore, this case does not yield a valid solution for x.

Thus, the only valid solution we have is $x = -1$. This is a crucial point in solving exponential equations: not all solutions obtained from the transformed equation will be valid solutions for the original equation. It is essential to check each solution to ensure it satisfies the original equation and the properties of exponential functions.

Verifying the Solution

We have found a potential solution for the exponential equation, $x = -1$. However, it's crucial to verify this solution to ensure it satisfies the original equation and that no errors were introduced during the solving process. Let's substitute $x = -1$ back into the original equation:

$(2^{2x+1})(3^{2x+2}) + 2^x(3^{x+2}) - 2 = 0$

Substituting $x = -1$, we get:

$(2^{2(-1)+1})(3^{2(-1)+2}) + 2^{-1}(3^{-1+2}) - 2 = 0$

Simplifying the exponents:

$(2^{-2+1})(3^{-2+2}) + 2^{-1}(3^{1}) - 2 = 0$

$(2^{-1})(3^{0}) + 2^{-1}(3) - 2 = 0$

Recall that any non-zero number raised to the power of 0 is 1. So, $3^0 = 1$. Also, 2^{-1} = rac{1}{2}. Substituting these values, we have:

(rac{1}{2})(1) + (rac{1}{2})(3) - 2 = 0

rac{1}{2} + rac{3}{2} - 2 = 0

Combining the fractions:

rac{4}{2} - 2 = 0

$2 - 2 = 0$

$0 = 0$

The equation holds true, which confirms that $x = -1$ is indeed a valid solution to the original exponential equation. This verification step is essential in mathematics to ensure the correctness of the solution, especially when dealing with equations that involve transformations and substitutions.

Conclusion

In conclusion, we have successfully solved the exponential equation $(2^{2x+1})(3^{2x+2}) + 2^x(3^{x+2}) - 2 = 0$. The solution process involved several key steps, each requiring a careful application of mathematical principles.

First, we simplified the equation by using the properties of exponents to rewrite the terms. This involved breaking down the exponential expressions and expressing them in a more manageable form. Then, we made a crucial substitution, letting $y = 6^x$, which transformed the exponential equation into a quadratic equation. This substitution allowed us to apply familiar algebraic techniques to solve for y.

Next, we solved the resulting quadratic equation, $18y^2 + 9y - 2 = 0$, by factoring. This yielded two possible values for y: rac{1}{6} and -rac{2}{3}. We then substituted these values back into the equation $y = 6^x$ to solve for x.

However, we encountered a critical point: one of the values of y led to an invalid solution for x because the exponential function $6^x$ cannot be negative. This highlighted the importance of verifying solutions in exponential equations. The only valid solution we found was $x = -1$.

Finally, we verified our solution by substituting $x = -1$ back into the original equation. This confirmed that our solution was correct and that no errors were made during the solving process. This comprehensive approach demonstrates the importance of not only finding a solution but also ensuring its validity. The techniques and strategies used in this article can be applied to a wide range of exponential equations, making it a valuable resource for anyone studying or working with exponential functions.