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Solving the Exponential Equation: (1/2)^x - 8 = 2^x
This article explores how to solve the exponential equation (1/2)^x - 8 = 2^x. We will utilize the properties of exponents and algebraic manipulation to arrive at the solution.
Understanding the Equation
The equation involves terms with different bases, (1/2) and 2, raised to the power of x. To solve this, we need to express both sides with the same base.
Solving the Equation
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Express (1/2)^x in terms of 2:
Recall that (1/2) can be written as 2^-1. Using the property of exponents, (a^m)^n = a^(m*n), we have:
(1/2)^x = (2^-1)^x = 2^(-x)
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Rewrite the equation:
Substitute 2^(-x) for (1/2)^x in the original equation:
2^(-x) - 8 = 2^x
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Isolate the exponential terms:
Add 8 to both sides:
2^(-x) = 2^x + 8
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Express all terms with the same base:
We need to express 8 as a power of 2. Since 8 = 2^3, we can rewrite the equation as:
2^(-x) = 2^x + 2^3
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Simplify and solve for x:
Now that all terms have the same base, we can focus on the exponents. To eliminate the negative exponent, multiply both sides by 2^x:
2^0 = 2^(2x) + 2^(x+3)
1 = 2^(2x) + 2^(x+3)
This equation is now in a form that resembles a quadratic equation. We can make a substitution to simplify it further: let y = 2^x.
The equation becomes:
1 = y^2 + 2^3 * y
1 = y^2 + 8y
Rearranging, we get:
y^2 + 8y - 1 = 0
This quadratic equation can be solved using the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / 2a
Where a = 1, b = 8, and c = -1.
Solving for y, we get:
y = (-8 ± √(8^2 - 4 * 1 * -1)) / 2 * 1
y = (-8 ± √(68)) / 2
y = (-8 ± 2√17) / 2
Therefore, y = -4 ± √17
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Solve for x:
Recall that y = 2^x. We need to solve for x in each of the solutions for y:
a. For y = -4 + √17:
2^x = -4 + √17
Since the base 2 is always positive, the right side cannot be negative. Therefore, this solution is not valid.
b. For y = -4 - √17:
2^x = -4 - √17
Similarly, the right side is negative, and this solution is not valid.
Conclusion
We found that the original equation (1/2)^x - 8 = 2^x has no real solutions. This means there is no real value for x that would satisfy the equation.