If 3x - 1 + 33 - x = 6, then what is 2x - 1 + 23 - x equal to ?
4
Understanding the Exponential Equation Problem
The problem asks us to find the value of an expression involving powers of 2, given an equation involving powers of 3. The equation is \(3^{x - 1} + 3^{3 - x} = 6\). We need to find the value of \(2^{x - 1} + 2^{3 - x}\). To solve this, we first need to find the value of 'x' from the given equation.
Solving the Equation for x
Let's solve the equation \(3^{x - 1} + 3^{3 - x} = 6\). This equation involves exponents with the variable 'x' in them. We can use substitution to make it simpler.
Let \(y = 3^{x - 1}\).
Now, let's look at the second term, \(3^{3 - x}\). We can rewrite this using exponent rules:
\(\qquad 3^{3 - x} = 3^{(2 + 1) - x} = 3^{2 - (x - 1)}\)
Using the rule \(a^{m-n} = a^m \cdot a^{-n} = a^m / a^n\), we get:
\(\qquad 3^{2 - (x - 1)} = 3^2 \cdot 3^{-(x - 1)} = 9 \cdot \frac{1}{3^{x - 1}}\)
Since we set \(y = 3^{x - 1}\), the second term \(3^{3-x}\) is equal to \(\frac{9}{y}\).
Now substitute 'y' into the original equation:
\(\qquad y + \frac{9}{y} = 6\)
To eliminate the fraction, multiply the entire equation by 'y' (we know \(y = 3^{x-1}\) must be positive, so \(y \neq 0\)):
\(\qquad y(y) + y\left(\frac{9}{y}\right) = 6y\)
\(\qquad y^2 + 9 = 6y\)
Rearrange this into a standard quadratic equation form \(ay^2 + by + c = 0\):
\(\qquad y^2 - 6y + 9 = 0\)
This quadratic equation looks familiar. It's a perfect square trinomial: \((y - 3)^2\).
\(\qquad (y - 3)^2 = 0\)
Taking the square root of both sides:
\(\qquad y - 3 = 0\)
Solving for 'y':
\(\qquad y = 3\)
Now, substitute back \(y = 3^{x - 1}\) to find 'x':
\(\qquad 3^{x - 1} = 3\)
Since the bases are the same, the exponents must be equal:
\(\qquad x - 1 = 1\)
Solving for 'x':
\(\qquad x = 1 + 1\)
\(\qquad x = 2\)
So, the value of 'x' that satisfies the given equation is 2.
Evaluating the Target Expression \(2^{x - 1} + 2^{3 - x}\)
Now that we have found \(x = 2\), we can substitute this value into the expression \(2^{x - 1} + 2^{3 - x}\) to find its value.
Substitute \(x=2\) into the expression:
\(\qquad 2^{x - 1} + 2^{3 - x} = 2^{2 - 1} + 2^{3 - 2}\)
Simplify the exponents:
\(\qquad = 2^1 + 2^1\)
Calculate the powers:
\(\qquad = 2 + 2\)
Perform the addition:
\(\qquad = 4\)
The value of the expression \(2^{x - 1} + 2^{3 - x}\) for \(x=2\) is 4.
Verifying the Solution for x=2
Let's quickly check if \(x=2\) works in the original equation \(3^{x - 1} + 3^{3 - x} = 6\):
Substitute \(x=2\):
\(\qquad 3^{2 - 1} + 3^{3 - 2} = 3^1 + 3^1 = 3 + 3 = 6\)
The equation holds true for \(x=2\). This confirms our value of 'x' is correct.
Summary of Steps
- Rewrite the given equation \(3^{x - 1} + 3^{3 - x} = 6\) using substitution.
- Let \(y = 3^{x - 1}\) and express \(3^{3 - x}\) in terms of 'y'.
- Formulate and solve the resulting quadratic equation for 'y'.
- Substitute back the value of 'y' to find the value of 'x'.
- Substitute the found value of 'x' into the expression \(2^{x - 1} + 2^{3 - x}\) and calculate the result.
The value of \(2^{x - 1} + 2^{3 - x}\) is 4.
Revision Table: Key Concepts
| Concept | Description | Example Used Here |
|---|---|---|
| Exponent Rule: \(a^{m-n}\) | \(a^{m-n} = \frac{a^m}{a^n}\) | \(3^{3-x} = 3^{2-(x-1)} = \frac{3^2}{3^{x-1}}\) |
| Exponent Rule: \(a^{-n}\) | \(a^{-n} = \frac{1}{a^n}\) | \(3^{-(x-1)} = \frac{1}{3^{x-1}}\) |
| Solving Exponential Equations | If \(a^m = a^n\) (and \(a > 0, a \neq 1\)), then \(m = n\). | \(3^{x-1} = 3^1 \implies x-1 = 1\) |
| Substitution | Replacing an expression with a single variable to simplify an equation. | Let \(y = 3^{x-1}\) |
| Quadratic Equation | An equation of the form \(ay^2 + by + c = 0\). | \(y^2 - 6y + 9 = 0\) |
| Perfect Square Trinomial | A trinomial that can be factored into \((py \pm q)^2\). | \(y^2 - 6y + 9 = (y-3)^2\) |
Additional Information on Exponents and Equations
Problems involving exponents often require using exponent rules to simplify expressions or equations. Recognizing patterns like perfect square trinomials in quadratic equations can significantly speed up finding solutions.
When solving equations with exponents, isolating the exponential term is often the first step. If different bases are involved, techniques like logarithms might be needed, but in this specific problem, a simple substitution and algebraic manipulation were sufficient because the exponents had a related structure (\(x-1\) and \(3-x\)). Notice that the exponents in the original equation \((x-1)\) and \((3-x)\) add up to \((x-1) + (3-x) = 2\). This specific structure is key to the substitution working out nicely into a standard quadratic form.
Similarly, the target expression \(2^{x-1} + 2^{3-x}\) has the same structure in its exponents. Once 'x' is found, evaluating the expression becomes a direct substitution and calculation.
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