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Magazine Chapter 1: Problem 32
Solve the equation for \(x\). \(3=4 \cdot 10^{-0.5 x}\)
Short Answer
Expert verified
x \approx 0.2498
Step by step solution
01
Isolate the Exponential Term
Start by isolating the exponential part of the equation. The original equation is \(3 = 4 \cdot 10^{-0.5x}\). To isolate \(10^{-0.5x}\), divide both sides of the equation by 4. This gives:\[ \frac{3}{4} = 10^{-0.5x} \]
02
Take the Logarithm of Both Sides
To solve for the exponent, take the logarithm of both sides of the equation. Use the base 10 logarithm:\[ \log_{10}\left(\frac{3}{4}\right) = \log_{10}\left(10^{-0.5x}\right) \]By the properties of logarithms, the right side can be simplified to:\[ \log_{10}\left(\frac{3}{4}\right) = -0.5x \]
03
Solve for x
Now, solve for \(x\) by dividing both sides of the equation by -0.5. This results in:\[ x = \frac{\log_{10}\left(\frac{3}{4}\right)}{-0.5} \]
04
Calculate x
Calculate \(x\) using a calculator. First, find \(\log_{10}\left(\frac{3}{4}\right)\):\[ \log_{10}\left(0.75\right) \approx -0.1249 \]Then, divide by -0.5:\[ x \approx \frac{-0.1249}{-0.5} = 0.2498 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Logarithmic Functions
Logarithmic functions are critical tools in mathematics, especially when you need to solve equations involving exponents. A logarithm essentially does the opposite of what an exponential function does. If you have an equation like \( b = a^x \), the logarithm helps you find the value of \( x \).
This becomes especially useful when tackling exponential equations, where the unknown variable is in the exponent.
Here are a few key points about logarithms:
This becomes especially useful when tackling exponential equations, where the unknown variable is in the exponent.
Here are a few key points about logarithms:
- Logarithmic Base: The base of the logarithm is the number that is raised to a power to get another number. In this problem, base 10 logarithms are used because the exponential equation involves powers of 10.
- Properties: One key property of logarithms used here is \( \log_b(b^x) = x \). This property lets us bring the exponent down, simplifying complex equations.
- Calculating logarithms: Most calculators have a logarithm function to help you find values like \( \log_{10}(3/4) \).
Solving Equations
Solving equations involves finding the value of an unknown variable that makes the equation true. In the exponential equation given, \( 3 = 4 \cdot 10^{-0.5 x} \), our task is to find the value of \( x \).
To solve such equations, follow these systematic steps:
To solve such equations, follow these systematic steps:
- Isolation of the Term: First, isolate the term that contains the unknown variable. Here, divide both sides by 4 to separate \( 10^{-0.5x} \) from the rest of the equation.
- Apply Logarithms: Taking the logarithm of both sides transforms the exponential equation into a linear form. This allows you to move the exponent down, converting the problem into one that's easier to solve.
- Calculate and Solve: Once simplified, the equation can be solved like any linear equation. Divide or multiply as needed to find \( x \).
Mathematical Problem-Solving
Mathematical problem-solving is an essential skill that involves understanding the problem, planning a strategy, carrying out the plan, and checking the results. The process we used to solve the equation \( 3 = 4 \cdot 10^{-0.5 x} \) is a classic example of this methodology.
Let's break down the problem-solving process:
Let's break down the problem-solving process:
- Understand: Clearly understand the equation and what needs to be solved. Identify the unknown variable and note the exponent as the place where the unknown resides.
- Plan: Develop a strategy to solve the problem. In this case, isolating the exponential term and then applying logarithms was our approach.
- Carry Out the Plan: Execute the steps, using mathematical operations such as division and logarithms to simplify and solve the equation.
- Check: Validate the solution by substituting \( x \) back into the original equation. This ensures the solution is correct and satisfies the initial problem.
