Chapter 5: Problem 64
Solve each equation. Use the change of base formula to approximate exact answers to the nearest hundredth when appropriate. $$4 \cdot 10^{2 x}+1=21$$
Short Answer
Expert verified
x ≈ 0.35
Step by step solution
01
Isolate the Exponential Expression
Start by isolating the term with the exponential expression. The equation is \(4 \cdot 10^{2x} + 1 = 21\). Subtract 1 from both sides to get: \(4 \cdot 10^{2x} = 20\).
02
Solve for the Exponential Term
Divide both sides by 4 to isolate the exponential term. Thus, \(10^{2x} = 5\).
03
Apply the Change of Base Formula
To solve \(10^{2x} = 5\), apply the logarithm to both sides. Using the change of base formula, \(2x = \log_{10} 5\). This is represented as \(2x = \frac{\ln 5}{\ln 10}\).
04
Solve for x
Simplify the expression by dividing both sides by 2. Thus, \(x = \frac{\ln 5}{2 \ln 10}\).
05
Substitute and Calculate
Calculate the value using a calculator. \(\ln 5 \approx 1.6094\) and \(\ln 10 \approx 2.3026\). Substitute these values in: \(x = \frac{1.6094}{2 \times 2.3026} \approx 0.349\). Therefore, the approximate solution is \(x \approx 0.35\) when rounded to the nearest hundredth.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Change of Base Formula
The change of base formula is a useful tool when dealing with exponential equations that require logarithmic solutions. This formula allows you to calculate logarithms in any base using a calculator that typically only offers natural log or common log functionalities. The formula is:
To use this formula effectively, follow these steps:
- \( \log_b a = \frac{\ln a}{\ln b} \)
To use this formula effectively, follow these steps:
- Identify the base you need to convert to a base of 10 or \( e \).
- Calculate the natural log of both \( a \) and \( b \).
- Divide \( \ln a \) by \( \ln b \), allowing you to find the equivalent logarithm in your desired base.
Logarithms
Logarithms are the inverse operations to exponentials. They help determine the exponent or power to which a base must be raised to obtain a specific value. In simpler terms, if \( b^x = a \), then \( x = \log_b a \).
Understanding the logarithmic functions is crucial in solving higher mathematics problems, especially those involving exponential growth or decay.
There are mainly two types of logarithms widely used:
For effective problem solving, always remember to:
Understanding the logarithmic functions is crucial in solving higher mathematics problems, especially those involving exponential growth or decay.
There are mainly two types of logarithms widely used:
- Natural Logarithms (denoted as \( \ln \), where the base is the constant \( e \))
- Common Logarithms (denoted as \( \log \), where the base is 10)
For effective problem solving, always remember to:
- Apply the same logarithmic operation to both sides of an equation.
- Simplify and reorganize to solve for the unknown variable.
Rounding Decimals
Rounding decimals is a fundamental math skill, crucial for approximating values accurately in calculations. In many mathematical and real-life applications, you are required to round numbers to a certain decimal place, whether it be for simplicity or because it closely aligns with measurable precision.
Here are some guiding tips to round numbers effectively:
For example, when calculating \( x = \frac{\ln 5}{2 \ln 10} \), the actual results can be quite lengthy. Rounding ensures that the solution not only fits within a practical number range but also matches the precision level required, in this case, the nearest hundredth. Hence, \( x \approx 0.35 \).
Here are some guiding tips to round numbers effectively:
- Identify the decimal place to which you need to round. Common places include the nearest tenth, hundredth, or thousandth.
- Look at the digit immediately to the right of your identified place. If it is 5 or greater, increase the identified digit by one. If it is less than 5, leave the digit unchanged.
For example, when calculating \( x = \frac{\ln 5}{2 \ln 10} \), the actual results can be quite lengthy. Rounding ensures that the solution not only fits within a practical number range but also matches the precision level required, in this case, the nearest hundredth. Hence, \( x \approx 0.35 \).