Solving the Equation ( x^{x_1 x_2 x_3} 24 ): A Step-by-Step Guide

Introduction

There are various methods to solve equations that may not appear straightforward at first glance. In this article, we will explore a detailed step-by-step process for solving the equation ( x^{x_1 x_2 x_3} 24 ). This problem will be broken down into several manageable parts to ensure clarity and comprehension. By the end of this guide, you will have a deep understanding of how to approach similar equations, particularly involving exponential and polynomial expressions.

Background and Preparation

Before diving into the solution, it is essential to understand the basics of the problem. The equation ( x^{x_1 x_2 x_3} 24 ) requires us to find the value(s) of ( x ) that satisfy this equation. This equation can be complex due to the presence of the exponent, which is a product of three variables ( x_1, x_2, ) and ( x_3 ).

Step 1: Simplify the Exponent

The exponent in the given equation needs to be simplified for easier manipulation. By introducing a new variable ( y ) such that ( y x - 3 ), we can rewrite the original equation in terms of ( y ).

Let's start by substituting ( y x - 3 ):

Since ( y x - 3 ), we have ( x y 3 ). Now, substituting ( x ) in the original equation:

( (y 3)^{y_1 y_2 y_3} 24 )

This step helps in transforming the equation into a more workable form. However, it also suggests that we need to find a simpler form of the exponent.

Step 2: Rewrite the Exponent

Next, we rewrite the left side of the equation in a simpler form:

( x^{x_1 x_2 x_3} x^2 (3x^2 - 3x 2) )

By letting ( y x^2 - 3x ), we can further simplify the equation:

( y^2 - 2y - 24 0 )

This quadratic equation will be solved to find the values of ( y ).

Step 3: Solve the Quadratic Equation

To solve the quadratic equation ( y^2 - 2y - 24 0 ), we use the quadratic formula:

( y frac{-b pm sqrt{b^2 - 4ac}}{2a} )

Here, ( a 1 ), ( b -2 ), and ( c -24 ). Plugging these values into the formula gives:

( y frac{2 pm sqrt{(-2)^2 - 4 cdot 1 cdot (-24)}}{2} )

Simplifying further:

( y frac{2 pm sqrt{4 96}}{2} )

( y frac{2 pm sqrt{100}}{2} )

( y frac{2 pm 10}{2} )

This yields two solutions for ( y ):

( y frac{12}{2} 6 ) and ( y frac{-8}{2} -4 )

Step 4: Substitute Back to Find ( x )

Now we substitute the values of ( y ) back to find ( x ).

For ( y 6 ):

Since ( y x^2 - 3x ), we have:

( x^2 - 3x - 6 0 )

The discriminant of this quadratic equation is:

( (-3)^2 - 4 cdot 1 cdot (-6) 9 24 33 )

Since the discriminant is positive, there are two real solutions:

( x frac{3 pm sqrt{33}}{2} )

However, these do not yield simple integer solutions.

For ( y -4 ):

Since ( y x^2 - 3x ), we have:

( x^2 - 3x - 4 0 )

Using the quadratic formula:

( x frac{3 pm sqrt{9 - 16}}{2} )

( x frac{3 pm sqrt{-7}}{2} )

Since the discriminant is negative, there are no real solutions from this case.

This leaves us with the simpler solutions:

For ( y 4 ):

Since ( y x^2 - 3x ), we have:

( x^2 - 3x - 4 0 )

Using the quadratic formula:

( x frac{3 pm 5}{2} )

( x 4 ) or ( x -1 )

For ( y -1 ):

Since ( y x^2 - 3x ), we have:

( x^2 - 3x 1 0 )

Using the quadratic formula:

( x frac{3 pm sqrt{9 - 4}}{2} )

( x frac{3 pm sqrt{5}}{2} )

This gives two potential solutions for ( x ):

( x frac{3 sqrt{5}}{2} ) or ( x frac{3 - sqrt{5}}{2} )

Summarizing, the real solutions to the original equation ( x^{x_1 x_2 x_3} 24 ) are:

( x 4, -4, 1, frac{3 sqrt{5}}{2}, frac{3 - sqrt{5}}{2} )

Conclusion

In conclusion, solving the equation ( x^{x_1 x_2 x_3} 24 ) involves a series of algebraic manipulations, including simplifying the exponent and solving quadratic equations. The solutions found are both real and complex, highlighting the variety of potential outcomes in such problems.