Problem 36 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 35-38: Use common logarithms to solve for \(x\) in terms of \(y\). $$ y=\frac{10^{x}-10^{-x}}{2} $$

Step-by-Step Solution

Answer

\(x = \log_{10}(y + \sqrt{y^2 + 1})\)

Step 1: Isolate Exponential Terms

To make the equation easier to solve, we first attempt to isolate the exponential terms. The given equation is \( y = \frac{10^x - 10^{-x}}{2} \). Multiply both sides by 2 to clear the fraction: \( 2y = 10^x - 10^{-x} \).

Step 2: Rewrite in Terms of Logarithms

We aim to express terms in a format that allows applying logarithms conveniently. Notice that 10\(^{-x}\) can be rewritten as \( \frac{1}{10^x} \). This gives us \( 2y = 10^x - \frac{1}{10^x} \). To facilitate taking logarithms, let's set \( a = 10^x \). Thus, \( 2y = a - \frac{1}{a} \).

Step 3: Clear Fraction by Multiplying

To easily eliminate the fraction in \( a - \frac{1}{a} \), multiply through by \( a \), giving us \( 2ya = a^2 - 1 \). Rearrange this as a quadratic equation: \( a^2 - 2ya - 1 = 0 \).

Step 4: Solve Quadratic for \(a\)

The above equation is a quadratic in \(a\). Solve it using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( b = -2y \), \( a = 1 \), and \( c = -1 \), so \( a = \frac{2y \pm \sqrt{(2y)^2 + 4}}{2} \).

Step 5: Apply Common Logarithm

Use the definition of logarithms to solve for \( x \). Since \( a = 10^x \), take the common logarithm: \( \log_{10}a = x \). Thus, \( x = \log_{10}\left( \frac{2y \pm \sqrt{4y^2 + 4}}{2} \right) \).

Step 6: Simplify the Expression

Simplify the term under the logarithm. Notice the expression inside the square root simplifies: \( \sqrt{4y^2 + 4} = \sqrt{4(y^2 + 1)} = 2\sqrt{y^2 + 1} \). Thus, \( x = \log_{10}\left(y + \sqrt{y^2 + 1} \right) \). The negative solution does not make sense in the context of common logarithms.

Key Concepts

Exponential Equations

Exponential equations are equations where the unknown variable appears as an exponent. In mathematical terms, if you have an equation like \( b^x = a \), this is an exponential equation. Solving these equations often involves isolating the term with the exponential expression on one side of the equation and then applying logarithms. Common logarithms, which are logarithms with base 10, are extremely useful in solving exponential equations where the base of the exponent is 10. Here’s a quick refresher:

• Isolate the exponent term, if possible.
• Take the logarithm of both sides of the equation.
• Solve for the variable.

In our exercise, we started with an exponential equation where terms like \(10^x\) were present. By strategically using algebraic manipulations and logarithms, we effectively solved for \( x \).

Quadratic Formula

The quadratic formula is a powerful tool used for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). The formula for finding the roots (solutions) of a quadratic equation is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here’s how it helps:

• Calculate the discriminant \( b^2 - 4ac \).
• Determine the number and type of solutions based on the discriminant.
• Use the formula to find the roots.

In the exercise, the logarithmic manipulation led to a quadratic in terms of \( a \), which was solved using this formula. This revealed the necessary expression for \( a \) which we eventually transformed back using logarithms to find \( x \).

Logarithmic Functions

Logarithmic functions are the inverse functions of exponential functions. If an exponential function is \( y = b^x \), then the corresponding logarithmic function is \( x = \log_b(y) \). For common logarithms, the base \( b \) is 10, noted as \( \log(y) \) or \( \log_{10}(y) \). Here’s what to keep in mind:

• They’re the reverse of exponentiation.
• Allow you to solve for the exponent in exponential equations.
• They satisfy the property \( \log_b(b^x) = x \).

By applying logarithmic functions strategically, you simplify complex expressions involving exponents into linear ones. In our solution, common logarithms were used to express \( x \) in terms of \( y \) after isolating exponential increments.

Solving Equations

Solving equations is the process of finding the unknown values that satisfy the given conditions of the equation. This can involve multiple steps: rearranging terms, factoring, using identities, and applying mathematical functions like logarithms.

• Identify what type of equation you have (linear, quadratic, exponential).
• Apply appropriate algebraic techniques.
• Verify solutions within the context of the problem.

In the original exercise, solving the equation required isolating terms, using quadratic formulas, and applying logarithmic properties. This methodical approach allowed us to express \( x \) in a simplified logarithmic form. Understanding which techniques to apply is crucial in converting a complex equation into a solvable form.