Problem 91 - Algebra and Trigonometry

Question

Each of these equations can be transformed into an equation of linear or quadratic type by applying the hint. Solve each equation. (a) \((x-1)^{\log (x-1)}=100(x-1)\) (b) \(\log _{2} x+\log _{4} x+\log _{8} x=11\) (c) \(4^{x}-2^{x+1}=3\)

Step-by-Step Solution

Answer

(a) x = 101; (b) x = 64; (c) x = \log_2 3.

Step 1: Simplify the equation (a)

First, observe that the equation \((x-1)^{\log(x-1)} = 100(x-1)\) can be simplified. By dividing both sides by \((x-1)\), assuming \(x-1 eq 0\): \[ (x-1)^{\log(x-1)-1} = 100 \]Let \(y = x-1\), then we have:\[ y^{\log(y) - 1} = 100 \]This equation implies \(\log(y) = 2\) because it simplifies into the quadratic-like form \(y^2 = 100,\) once considered under the exponential and logarithmic properties.

Step 2: Solve for y in equation (a)

Since \(\log(y) = 2\) implies \(y = 10^2 = 100,\) substitute back \(y = x - 1\) to find \(x\). This gives us:\[ x - 1 = 100 \]\[ x = 101 \]

Step 3: Solve equation (b)

Convert each term in the equation \(\log_2 x + \log_4 x + \log_8 x = 11\) using the change of base formula:\[ \log_2 x = a, \quad \log_4 x = \frac{a}{2}, \quad \log_8 x = \frac{a}{3} \]This transforms the equation to:\[ a + \frac{a}{2} + \frac{a}{3} = 11 \]Combine the terms over a common denominator:\[ 6a + 3a + 2a = 66 \]\[ 11a = 66 \]

Step 4: Solve for x in equation (b)

Solve for \(a\):\[ a = 6 \]Since \(a = \log_2 x = 6\), it implies:\[ x = 2^6 = 64 \]

Step 5: Transform equation (c)

For the equation \(4^x - 2^{x+1} = 3\), notice that \(4^x\) can be rewritten as \((2^2)^x = (2^x)^2\). Let \(y = 2^x\); then the equation becomes:\[ y^2 - 2y = 3 \]

Step 6: Solve the quadratic equation (c)

Reorganize this into the standard quadratic form:\[ y^2 - 2y - 3 = 0 \]Factoring the quadratic gives:\[ (y - 3)(y + 1) = 0 \]This means the solutions for \(y\) are \(y = 3\) or \(y = -1\). However, since \(y = 2^x\) must be positive, discard \(y = -1\). Hence, \(y = 3\).

Step 7: Solve for x in equation (c)

If \(y = 3\) and \(y = 2^x\), then:\[ 2^x = 3 \]Taking the logarithm of both sides gives us:\[ x = \log_2 3 \]

Key Concepts

Logarithmic Equations

Logarithmic equations involve variables located within a logarithm. Solving them usually requires using logarithmic properties and transforming them into a different form. Let's break down the process.

• Convert logs to a common base: Consider the transformation using the change of base formula. For example, for the terms \(\log_2 x\), \(\log_4 x\), and \(\log_8 x\), convert them to a common base, say base 2.
• Add like terms: Once transformed, add the terms. For instance, \(a + \frac{a}{2} + \frac{a}{3} = 11\) can be solved by finding a common denominator.
• Solve for the variable: After transforming and simplifying, solve for the variable normally, such as finding \(x\) from \(2^x = \text{some value}\).

Using these steps supports solving logarithmic equations effectively. The key is the conversion and manipulation of log expressions, which can sometimes turn complex equations into simpler arithmetic calculations.

Quadratic Equations

Quadratic Equations are equations of the form \(ax^2 + bx + c = 0\). They can be solved by a variety of methods including factoring, completing the square, and using the quadratic formula. In our context, they emerged indirectly in transformed forms like \(y^2 - 2y - 3 = 0\).

• Recognize the structure: Understand when an equation can be written in a quadratic form. It might not always be explicit, so identifying patterns can help. For example, converting \(4^x = (2^x)^2\) is key.
• Factor the equation: If it factors neatly like \((y-3)(y+1) = 0\), find the solutions from the factors.
• Validate solutions: Ensure they fit the original variable's conditions. For instance, discard \(y = -1\) since \(2^x\) cannot be negative.

Quadratic equations are foundational in algebra and appear frequently across varied contexts, even where they aren't obvious at first glance.

Exponential Functions

Exponential functions take the form \(a^x\), where \(a\) is a constant, and \(x\) is the exponent. They are pivotal when variables appear as exponents.

• Transform for simplification: Convert expressions like \(4^x\) into \((2^2)^x\). Such transformations simplify the algebraic manipulation.
• Utilize logarithms for solving: If stuck with an expression like \(2^x = 3\), apply logarithms to both sides to isolate \(x\). This shifts the problem to a simpler algebraic domain.
• Understand growth and decay: Exponential functions explain rapid growth or decay, which is relevant in real-world applications beyond simple algebra problems.

Grasping exponential functions is crucial for advanced studies in mathematics, as they model numerous natural phenomena and statistical growth rates.