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Magazine Chapter 4: Problem 82
Solve each equation. Round answers to four decimal places. $$\left(1+\frac{r}{4}\right)^{20}=3$$
Short Answer
Expert verified
r 鈮� 0.0511
Step by step solution
01
Isolate the expression with the variable
The given equation is \ \ \ \( \left(1+\frac{r}{4}\right)^{20}=3 \). The expression \ \( \left(1+\frac{r}{4}\right) \) is already isolated.
02
Apply the logarithm
Apply the natural logarithm (ln) to both sides to simplify the exponent: \ \ \ \( \ln\left( \left(1+\frac{r}{4}\right)^{20} \right) = \ln(3) \).
03
Use log properties
Utilize the logarithmic property \ \( \ln(a^b) = b\ln(a) \) to move the exponent to the front: \ \ \ \( 20 \ln\left(1+\frac{r}{4}\right) = \ln(3) \).
04
Solve for the logarithm
Divide both sides of the equation by 20: \ \ \ \( \ln\left(1+\frac{r}{4}\right) = \frac{\ln(3)}{20} \).
05
Exponentiate both sides
Exponentiate both sides using the base \ e \ to get rid of the natural logarithm: \ \ \ \( 1+\frac{r}{4} = \exp\left(\frac{\ln(3)}{20}\right) \).
06
Solve for r
Isolate \ r \ by performing the following steps: \ \ \ 1. Subtract 1 from both sides: \ \( \frac{r}{4} = \exp\left(\frac{\ln(3)}{20}\right) - 1 \). \ \ 2. Multiply both sides by 4: \ \( r = 4 \left( \exp\left(\frac{\ln(3)}{20}\right) - 1 \right) \).
07
Calculate the numerical value
Compute the numerical value using a calculator while rounding to four decimal places: \ \ \ \( r \approx 4 \left( \exp\left(\frac{\ln(3)}{20}\right) - 1 \right) \approx 0.0511 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Natural Logarithm
The natural logarithm, often written as \(\text{ln}\), is a logarithm with the base \(e\). Here, \(e\) is a mathematical constant approximately equal to 2.71828. The natural logarithm is widely used in mathematics, especially in situations involving growth and decay problems. For example, in our exercise, we used the natural logarithm to help solve the equation involving an exponent.
Applying the natural logarithm to both sides of an equation allows us to use logarithmic properties to simplify the equation. This approach is particularly helpful when dealing with exponents. By converting the exponential equation into a logarithmic form, we can make the variable-containing expression easier to manage and solve.
Applying the natural logarithm to both sides of an equation allows us to use logarithmic properties to simplify the equation. This approach is particularly helpful when dealing with exponents. By converting the exponential equation into a logarithmic form, we can make the variable-containing expression easier to manage and solve.
Logarithmic Properties
Logarithmic properties are rules that describe how logarithms behave. These properties are essential when solving logarithmic and exponential equations. One key property used in the solution is the power rule: \( \text{ln}(a^b) = b \text{ln}(a) \). This rule allows us to move the exponent to the front of the logarithm, simplifying our equation significantly.
Another important property is the product rule, which states \( \text{ln}(ab) = \text{ln}(a) + \text{ln}(b) \). There's also the quotient rule, which says \( \text{ln}\frac{a}{b} = \text{ln}(a) - \text{ln}(b) \). Understanding these logarithmic properties is crucial for both simplifying complex equations and solving them efficiently. In our exercise, we used the power rule to handle the exponent by converting the equation into a simpler form that we could further solve.
Another important property is the product rule, which states \( \text{ln}(ab) = \text{ln}(a) + \text{ln}(b) \). There's also the quotient rule, which says \( \text{ln}\frac{a}{b} = \text{ln}(a) - \text{ln}(b) \). Understanding these logarithmic properties is crucial for both simplifying complex equations and solving them efficiently. In our exercise, we used the power rule to handle the exponent by converting the equation into a simpler form that we could further solve.
Exponentiation
Exponentiation refers to the process of raising a number to a power. In the context of solving equations, exponentiation is often reversed using logarithms, and vice versa. For instance, if we have \(a^b = c\), we might apply the natural logarithm to both sides to simplify the equation: \( \text{ln}(a^b) = \text{ln}(c) \).
Once the equation is in logarithmic form, we can use logarithmic properties to work with the exponent more easily. After simplifying and solving for the variable, as seen in our solution, we often need to exponentiate again to revert back from the logarithmic form. In the final step of our exercise, we exponentiated both sides using the base \(e\), undoing the natural logarithm and solving for the variable. This back-and-forth between modes鈥攁pplying logarithms and exponentiating鈥攊s a common practice in solving exponential equations.
Once the equation is in logarithmic form, we can use logarithmic properties to work with the exponent more easily. After simplifying and solving for the variable, as seen in our solution, we often need to exponentiate again to revert back from the logarithmic form. In the final step of our exercise, we exponentiated both sides using the base \(e\), undoing the natural logarithm and solving for the variable. This back-and-forth between modes鈥攁pplying logarithms and exponentiating鈥攊s a common practice in solving exponential equations.
