Solving Exponential Equations Through Quadratic Form and Logarithms

In mathematics, solving exponential equations can often be approached using techniques such as converting the equation to a quadratic form, utilizing logarithms, and factoring. Let's explore a specific problem where these methods are applied effectively.

Problem Statement

Consider the following equation: 2^x 1 - 3 2^{2x} - 5. Our goal is to find the value of x.

Approach: Quadratic Form and Simultaneous Equations

First, we rearrange the equation to show the equality between both sides:

2^x 1 - 3 2^{2x} - 5

We can simplify this to:

2^x 1 - 3 2^{2x} - 5

Further simplification by adding 3 to both sides gives us:

2^x 1 2^{2x} - 2

Recognize that 2^3 8, so we rewrite the equation as:

2^x 1 2^{2x} - 2^3

If we subtract both sides by 2^x and add 2 to both sides, we get:

1 2^{2x} - 2^x - 2^3 2^1

Further simplifies to:

1 2^{2x} - 2^x - 8 2^1

By guessing and checking or further manipulation, we can determine that:

2^{2x} - 2^x - 7 0

Solving the Quadratic Equation

Let's denote y 2^x, and put this into the equation:

y^2 - y - 8 0

This is a standard quadratic equation of the form ay^2 by c 0. We can solve this using the quadratic formula:

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

Substituting the values, we get:

y frac{1 pm sqrt{1 32}}{2} frac{1 pm sqrt{33}}{2}

The positive solution for y is:

y frac{1 sqrt{33}}{2}

Back to x

Since y 2^x, we can take the logarithm of both sides to solve for x:

2^x frac{1 sqrt{33}}{2}

x log_2 left( frac{1 sqrt{33}}{2} right)

Alternative Method: Quadratic Form

Alternatively, we can directly solve the given quadratic form:

2^{2x} - 2 cdot 2^x - 8 0

By guess and check, we find that:

x 4

Substituting x 4 back into the original equation to verify:

2^x 1 - 3 2^{2x} - 5

2^4 1 - 3 2^{2cdot4} - 5

16 1 - 3 2^8 - 5

14 256 - 5

14 251

Complex Solutions

Note that when dealing with quadratic forms and logarithms, complex solutions may arise. Using MATLAB or other computational tools can help verify the roots:

roots([1, -2, -8])

This gives:

4, -2

Further solving:

log(4) / log(2) 2

log(-2) / log(2) 1 icdot4.5324

These solutions are complex, but only the real solution is typically relevant.

Conclusion

The primary solution for the given equation is x 4. However, understanding the process and potential complex solutions is crucial in more advanced mathematical contexts.