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Magazine Chapter 4: Problem 37
Use common logarithms to solve for \(x\) in terms of \(y\) $$y=\frac{10^{x}-10^{-x}}{10^{x}+10^{-x}}$$
Short Answer
Expert verified
\(x = \frac{1}{2} \log_{10}\left(\frac{-1-y}{1-y}\right)\).\)
Step by step solution
01
Identify the Equation
The given equation is \(y = \frac{10^{x} - 10^{-x}}{10^{x} + 10^{-x}}\). This needs to be simplified using common logarithms to solve for \(x\) in terms of \(y\).
02
Multiply Numerator and Denominator by \(10^x\)
To simplify, multiply both the numerator and the denominator by \(10^x\) to remove the negative exponent: \[y = \frac{10^{x}\cdot 10^x - 1}{10^{x}\cdot 10^x + 1} = \frac{10^{2x} - 1}{10^{2x} + 1}\].
03
Rearrange the Equation
Rearrange the equation by expressing \(10^{2x}\) in terms of \(y\): \[y(10^{2x} + 1) = 10^{2x} - 1\]. Expand and then rearrange: \[y \cdot 10^{2x} + y = 10^{2x} - 1 \].
04
Isolate \(10^{2x}\)
Bring all the terms with \(10^{2x}\) to one side: \[10^{2x} - y \cdot 10^{2x} = -1 - y\]. Factor out \(10^{2x}\): \[10^{2x}(1 - y) = -1 - y\].
05
Solve for \(10^{2x}\)
Solve for \(10^{2x}\) by dividing both sides by \((1 - y)\): \[10^{2x} = \frac{-1-y}{1-y}\].
06
Apply Logarithms to Solve for \(x\)
To solve for \(x\), take the logarithm base 10 of both sides: \[2x = \log_{10}\left(\frac{-1-y}{1-y}\right)\]. Finally, solve for \(x\) by dividing by 2: \[x = \frac{1}{2} \log_{10}\left(\frac{-1-y}{1-y}\right)\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Exponential Equations
Exponential equations are mathematical expressions where a variable appears in the exponent. They take the form of expressions like \(a^x = b\), where \(a\) is a constant base, and \(x\) is the variable we want to solve for.
To solve exponential equations, especially when they involve complex expressions, we often use properties of logarithms. Recognizing these kinds of equations is crucial because the power they represent grows or shrinks very quickly.
In the exercise, the expression \(10^{x}\) is an example of an exponential function. Also, notice the challenge that arises when we have negative exponents, such as \(10^{-x}\). In such cases, we multiply through by the base to simplify. This clean-up process makes it easier to isolate the exponential term, opening the way to apply logarithms for solving.
To solve exponential equations, especially when they involve complex expressions, we often use properties of logarithms. Recognizing these kinds of equations is crucial because the power they represent grows or shrinks very quickly.
In the exercise, the expression \(10^{x}\) is an example of an exponential function. Also, notice the challenge that arises when we have negative exponents, such as \(10^{-x}\). In such cases, we multiply through by the base to simplify. This clean-up process makes it easier to isolate the exponential term, opening the way to apply logarithms for solving.
Diving into Logarithmic Equations
Logarithmic equations involve the logarithm of an expression. They're the inverse of exponential equations and allow us to solve for exponents. For instance, the expression \(x = \log_a b\) translates to \(a^x = b\).
In solving equations like the one in the exercise, logarithms become important once we've isolated the exponential part. By applying logarithms, we transform the exponential equation into a form where it's easier to solve for the variable.
Here, we used the common logarithm, which is a logarithm with base 10, written as \(\log_{10}\). The utility of the common logarithm lies in its widespread applicability in practical problems involving base 10, such as those in science and engineering. In the solution, common logarithms helped us rewrite \(10^{2x}\) in terms of a familiar format to solve for \(x\).
In solving equations like the one in the exercise, logarithms become important once we've isolated the exponential part. By applying logarithms, we transform the exponential equation into a form where it's easier to solve for the variable.
Here, we used the common logarithm, which is a logarithm with base 10, written as \(\log_{10}\). The utility of the common logarithm lies in its widespread applicability in practical problems involving base 10, such as those in science and engineering. In the solution, common logarithms helped us rewrite \(10^{2x}\) in terms of a familiar format to solve for \(x\).
Strategies for Solving Equations
Solving equations, whether they鈥檙e exponential or logarithmic, involves strategic steps that simplify the process. The main goal is to isolate the variable of interest. In the presented exercise, the variable to solve for is \(x\).
Let's recap the solving process used in the solution:
Let's recap the solving process used in the solution:
- Simplify the equation by removing negative exponents. In the exercise, multiplying by \(10^x\) helped eliminate \(10^{-x}\).
- Rearrange terms to break down the equation. This was seen when we expressed everything in terms of \(10^{2x}\).
- Isolate the exponential expression on one side. This creates an equation that can be tackled with logarithms.
- Apply logarithms to resolve the exponential part, making it possible to then solve for the variable directly.
- Finally, simplify the equation if necessary to get the solution in a form that is easy to interpret.
