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Magazine Chapter 4: Problem 120
Solve the equation. \((\log x)^{2}=\log x^{3}\)
Short Answer
Expert verified
The solutions are \(x = 1\) and \(x = 1000\).
Step by step solution
01
Understand the Equation
The given equation is \((\log x)^{2}=\log x^{3}\). Our goal is to solve for \(x\).
02
Simplify the Right Side
Use the logarithm power rule \(\log a^{b}=b \log a\). Therefore, the equation becomes \((\log x)^{2}=3 \log x\).
03
Set up a Quadratic Equation
Let \(y = \log x\). Then the equation \(y^{2}=3y\) can be written.
04
Solve the Quadratic Equation
Rearrange the equation to \(y^{2} - 3y = 0\). Factor the quadratic equation: \y(y - 3) = 0\.
05
Find the Solutions for \(y\)
Set each factor equal to zero: \y = 0\ or \y = 3\. This gives the solutions \(\log x = 0\) or \(\log x = 3\).
06
Solve for \(x\)
Use the fact that \(\log x = 0\) implies \(x = 10^0 = 1\), and \(\log x = 3\) implies \(x = 10^3 = 1000\).
07
Verify the Solutions
Check both solutions in the original equation: For \(x = 1\), \((\log 1)^{2} = 0 = \log 1^{3}\), which is true. For \(x = 1000\), \((\log 1000)^{2} = 9 = \log 1000^{3}\), which is also true.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Logarithms
Logarithms help us solve equations involving exponents. They are the inverse operation of exponentiation.
For instance, if we have an equation like \(b^y = x\), the logarithmic form is \(\text{log}_b(x) = y\). In this problem, \(\text{log}(x)\) is treated with its properties.
Using the power rule in logarithms, \(\text{log}(a^b) = b \text{log}(a)\), simplifies complex log equations.
This allows converting \( (\text{log}(x))^2 = \text{log}(x^3) \) to \( (\text{log}(x))^2 = 3 \text{log}(x)\). This step simplifies abstract expressions into more familiar algebraic terms.
For instance, if we have an equation like \(b^y = x\), the logarithmic form is \(\text{log}_b(x) = y\). In this problem, \(\text{log}(x)\) is treated with its properties.
Using the power rule in logarithms, \(\text{log}(a^b) = b \text{log}(a)\), simplifies complex log equations.
This allows converting \( (\text{log}(x))^2 = \text{log}(x^3) \) to \( (\text{log}(x))^2 = 3 \text{log}(x)\). This step simplifies abstract expressions into more familiar algebraic terms.
Exponents
Exponents signify repeated multiplication.
For example, \(x^3 = x \times x \times x\).
In logarithms, they play a critical role in simplifying expressions thanks to the power rule.
In our exercise, converting \( \text{log}(x^3)\) to \(3 \text{log}(x)\) shows this interaction.
By mastering exponent rules, complex logs can be broken down, making equations more manageable.
When faced with \( \text{log}(x) \) terms raised to a power, always look to use exponent rules to simplify.
For example, \(x^3 = x \times x \times x\).
In logarithms, they play a critical role in simplifying expressions thanks to the power rule.
In our exercise, converting \( \text{log}(x^3)\) to \(3 \text{log}(x)\) shows this interaction.
By mastering exponent rules, complex logs can be broken down, making equations more manageable.
When faced with \( \text{log}(x) \) terms raised to a power, always look to use exponent rules to simplify.
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying equations. It's like solving a puzzle where you move pieces around.
In the exercise, substitution is used to make the equation more familiar.
Letting \log(x) = y\turns \( (\text{log}(x))^2 = 3 \text{log}(x) \) into \( y^2 = 3y \).
This opens up familiar quadratic solution methods.
By mastering substitution and rearrangement, complex problems become approachable. The aim is always to simplify the structure of an equation.
In the exercise, substitution is used to make the equation more familiar.
Letting \log(x) = y\turns \( (\text{log}(x))^2 = 3 \text{log}(x) \) into \( y^2 = 3y \).
This opens up familiar quadratic solution methods.
By mastering substitution and rearrangement, complex problems become approachable. The aim is always to simplify the structure of an equation.
Factoring
Factoring is a method to solve polynomial equations.
For quadratic equations, it breaks down expressions into products of simpler terms.
In our exercise, \ y^2 - 3y = 0\ becomes \( y(y-3) = 0\).
Once factored, setting each term to zero leads us to the solutions.
In this case, finding \(y = 0 \) or \( y = 3 \). This step converts polynomial solutions into linear ones.
Understanding factoring is essential, as it reveals potential solutions hiding in complex polynomial forms.
For quadratic equations, it breaks down expressions into products of simpler terms.
In our exercise, \ y^2 - 3y = 0\ becomes \( y(y-3) = 0\).
Once factored, setting each term to zero leads us to the solutions.
In this case, finding \(y = 0 \) or \( y = 3 \). This step converts polynomial solutions into linear ones.
Understanding factoring is essential, as it reveals potential solutions hiding in complex polynomial forms.
