Problem 35 - 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 the Exponential Expressions

We start with the equation \( y = \frac{10^{x} + 10^{-x}}{2} \). To eliminate the fraction, multiply both sides by 2 to get \( 2y = 10^x + 10^{-x} \).

Step 2: Substitute for the Exponential Expression

Recognize that \( 10^{-x} = \frac{1}{10^x} \), so rewrite the expression: \( 2y = 10^x + \frac{1}{10^x} \).

Step 3: Express in Terms of a Single Exponential

Multiply through by \(10^x\) to eliminate the fraction: \((2y)10^x = (10^x)^2 + 1\). Simplify to obtain the quadratic equation: \((10^x)^2 - (2y)10^x + 1 = 0\).

Step 4: Solve the Quadratic Equation

Apply the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -(2y), c = 1 \). Thus, \( t = \frac{2y \pm \sqrt{(2y)^2 - 4}}{2} \).

Step 5: Substitute Back for Exponential Expression

Recall that \( t = 10^x \), so \( 10^x = \frac{2y+\sqrt{4y^2-4}}{2} = y + \sqrt{y^2 - 1} \).

Step 6: Solve for x Using Logarithms

Apply the common logarithm to both sides: \( x = \log_{10}(y + \sqrt{y^2-1}) \).

Key Concepts

Exponential Equations

An exponential equation is one in which the variables appear as exponents. Such equations are quite common in scientific calculations and growth models. In the exercise, we started with the expression involving exponential terms, specifically with base 10.
Step 1 of the solution described isolating these exponential parts, which is crucial. This lets us address the components that have a variable in the exponent, like the terms in the form of

• \( 10^x \)
• \( 10^{-x} \)

The primary goal was to manipulate the equation until these terms become evident, by removing fractions and combining similar terms.

It's important to remember that dealing with exponential equations often involves rewriting expressions to highlight the exponential contributions. Once isolated, they can sometimes be turned into quadratic forms, allowing us to use methods from algebra to find solutions.

Quadratic Equations

Quadratic equations are polynomial equations of degree 2. They typically have a standard form:
\( ax^2 + bx + c = 0 \). In our exercise, step 3 transformed the original equation into a quadratic form by multiplying through by \( 10^x \), which was essential to solve for \( x \).

Once in the quadratic form:

• \((10^x)^2 - (2y)10^x + 1 = 0\)

we applied the quadratic formula, a reliable method to find solutions for quadratic equations. The quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) helps us find the roots of the equation:

• \( a = 1 \)
• \( b = -2y \)
• \( c = 1 \)

Recognizing the quadratic structure allowed us to isolate the exponential expression \( t = 10^x \) in the problem, facilitating the next steps in solving for \( x \).

Logarithmic Functions

Logarithms provide a way to work with exponential equations by simplifying the manipulation of powers.

In the final part of the problem, we leveraged logarithms to solve for \( x \). Common logarithms have a base of 10 and are denoted as \( \log_{10} \). Applying the logarithm to both sides of the equation introduces the ability to handle the power inside logarithms more conveniently:

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

This uses the property that if \( 10^x = a \), then \( x = \log_{10}(a) \). It directly reverses the process of exponentiation, providing a solution for \( x \).

Understanding how to use logarithmic functions is crucial for solving equations where the variable is an exponent, especially when traditional algebraic techniques are insufficient on their own.