Problem 53 - College Algebra with Modeling and Visualization

Question

Solve each equation. Use the change of base formula to approximate exact answers to the nearest hundredth when appropriate. (a) \(10^{x}=0.01 (b) \)10^{x}=7\( (c) \)10^{x}=-4$

Step-by-Step Solution

Answer

(a) -2, (b) 0.85, (c) No solution

Step 1: Solve Equation (a)

The equation given is \(10^x = 0.01\). To solve this, take the logarithm of both sides using base 10:\[\log_{10}(10^x) = \log_{10}(0.01) \]By the properties of logarithms, \(\log_{10}(10^x) = x\), so:\[x = \log_{10}(0.01)\]Since 0.01 is equivalent to \(10^{-2}\), we have:\[x = -2\]Thus, \(x = -2\).

Step 2: Solve Equation (b)

The equation is \(10^x = 7\). Again, take the logarithm of both sides using base 10:\[\log_{10}(10^x) = \log_{10}(7) \]This simplifies to \(x = \log_{10}(7)\). Use a calculator to find this:\[x \approx 0.845\]Thus, \(x \approx 0.85\), rounded to the nearest hundredth.

Step 3: Solve Equation (c)

The equation is \(10^x = -4\). Since the base 10 raised to any real number will never be negative, this equation has no real solutions.Thus, there is no solution for this equation in the real number system.

Key Concepts

Change of Base Formula

One of the useful tools when dealing with logarithms is the change of base formula. This formula is important when you're trying to calculate logarithms on calculators that offer limited base options, commonly base 10 (common log) and base \(e\) (natural log). The change of base formula states:

• \( \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \)

where \(c\) is any positive number different from 1, typically 10 or \(e\). This formula allows you to convert a logarithm from one base to another, making calculations possible even when direct computation isn't feasible. This change of base is particularly helpful in solving complex equations or when needing approximations, as it provides a way to use more accessible calculation platforms. For example, if you had \( \log_2(8) \), using the change of base formula, it would convert to \( \frac{\log_{10}(8)}{\log_{10}(2)} \) or \( \frac{\ln(8)}{\ln(2)} \), making it possible to solve using common calculator functions.

Exponential Equations

An exponential equation is a type of equation in which a variable appears as an exponent. These kinds of equations often arise in scenarios involving growth or decay, such as population studies, radioactive decay, or compound interest. The general form looks like \( a^x = b \), where \(a\) is the base and \(b\) is a constant term.

To solve an exponential equation, a common method includes taking the logarithm of both sides, which helps "bring down" the exponent, making the equation easier to manage. For example, to solve for \(x\) in \(10^x = 7\), you take:

• \( \log_{10}(10^x) = \log_{10}(7) \)

This simplifies to \(x = \log_{10}(7)\). By calculating \( \log_{10}(7) \), we can approximate our answer to the nearest hundredth, which is approximately \(0.85\). This technique is critical for finding solutions to exponential equations that aren't as straightforward as simple powers.

Properties of Logarithms

Understanding the properties of logarithms is essential for manipulating and solving equations that involve these operations.

• Product Property: \( \log_b(MN) = \log_b(M) + \log_b(N) \)
• Quotient Property: \( \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \)
• Power Property: \( \log_b(M^p) = p \cdot \log_b(M) \)

These properties help in simplifying complex logarithmic expressions and solving equations. For instance, knowing that \( \log_{10}(10^x) = x \) directly aids in solving exponential equations by simplifying them directly to a linear term. This becomes crucial when you work through equations such as \(10^x = 0.01\) in converting it to the simpler form of \(-2\) using the power property that acknowledges \( \log_{10}(10^{-2}) = -2 \). By mastering these properties, you can confidently approach a broad range of mathematical problems involving logarithms.