Problem 9 - Calculus

Question

Solve for \(x\) using logs. $$20=50(1.04)^{x}$$

Step-by-Step Solution

Answer

x ≈ -19.28

Step 1: Isolate the Exponent

First, divide both sides of the equation by 50 to remove the coefficient from the exponential term. \[\frac{20}{50} = (1.04)^{x}\]This simplifies to:\[0.4 = (1.04)^{x}\]

Step 2: Apply Logarithms to Both Sides

Take the logarithm of both sides to solve for \(x\). Using the natural logarithm (ln) or common logarithm (log) is standard.\[\log(0.4) = \log((1.04)^{x})\]According to the properties of logarithms, this becomes:\[\log(0.4) = x \cdot \log(1.04)\]

Step 3: Solve for x

Now, solve for \(x\) by dividing both sides by \(\log(1.04)\):\[x = \frac{\log(0.4)}{\log(1.04)}\]Use a calculator to evaluate the logarithms and find the value of \(x\).

Key Concepts

Exponential Equations

Exponential equations are equations where the variable is in the exponent. These equations typically have the form \(a=b \cdot c^{x}\), where you need to solve for \(x\). The key feature of solving exponential equations is isolating the exponential term. This makes it easier to work with the equation and apply methods like logarithms.To start solving an exponential equation, you often divide both sides by the coefficient of the exponential term. This step simplifies the problem and isolates the base and the exponent. From there, you can use different mathematical techniques to solve for the exponent.

Properties of Logarithms

Logarithms are powerful mathematical tools used to simplify exponential terms. Understanding the properties of logarithms can make solving equations much easier. Here are a few useful properties:

• Product Property: \(\log(ab) = \log(a) + \log(b)\)
• Quotient Property: \(\log\left(\frac{a}{b}\right) = \log(a) - \log(b)\)
• Power Property: \(\log(a^{b}) = b \cdot \log(a)\)

In the context of solving exponential equations, the Power Property is particularly useful. It allows you to bring the exponent down in front of the log, transforming the equation into a simpler, linear form. This is a crucial step in finding the value of the unknown exponent, \(x\). More importantly, understanding these properties enables you to apply logarithms properly when manipulating equations.

Solving Equations with Logs

To solve equations with logs, you first isolate the exponential part of the equation. Once that's done, you can apply a logarithm to both sides of the equation. This is necessary because it allows you to use the Power Property of logarithms to bring the variable exponent down. Here’s a step-by-step of how it applies:1. **Take the Logarithm of Both Sides**: This can be either a common logarithm (base 10) or a natural logarithm (base \(e\)).2. **Apply the Power Property**: Once you have \(\log((1.04)^{x})\), rewrite it as \(x \cdot \log(1.04)\).3. **Solve for the Variable**: After the exponent is down, you can solve for \(x\) by dividing both sides by \(\log(1.04)\).Use a calculator to compute the final value. This process turns a potentially complex exponential equation into a simpler linear equation that is straightforward to solve. With practice, solving equations with logs becomes an invaluable skill in mathematics.