Problem 75
Question
Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ \log x^{2}=2 $$
Step-by-Step Solution
Verified Answer
The exact solution is \( x = 10 \). No approximation needed.
1Step 1: Use the Property of Logarithms
The given equation is \( \log x^2 = 2 \). We can use the property of logarithms \( \log a^b = b \log a \) to rewrite the equation. Here, this becomes \( 2 \log x = 2 \).
2Step 2: Isolate the Logarithmic Term
Divide both sides of the equation by 2 to isolate the logarithm: \( \log x = 1 \).
3Step 3: Solve for x
To solve \( \log x = 1 \), we can rewrite it in exponential form: \( x = 10^1 \), so \( x = 10 \).
4Step 4: Verify the Solution
Substitute \( x = 10 \) back into the original equation to verify: \( \log 10^2 = 2 \). Since \( 10^2 = 100 \), and \( \log 100 = 2 \), the solution is verified. Thus, \( x = 10 \) is correct.
Key Concepts
Property of LogarithmsExponential FormVerification of Solutions
Property of Logarithms
The property of logarithms is a handy rule that helps simplify expressions involving logs. Specifically, one essential property that we used in solving the equation \( \log x^2 = 2 \) is the power rule. This rule states that \( \log a^b = b \log a \). It means you can take the exponent out of the logarithm and multiply it by the logarithm of the base itself.
For example, applying this to our problem, we started with \( \log x^2 \). Through the property of logarithms, we transformed it to \( 2 \log x \). This transformation simplifies the equation and makes it easier to solve the problem step by step.
Using these properties allows us to work with simplified expressions. When the exponents in the logarithmic function appear, rely on this property to move them outside the "log" for easier manipulation. These properties expand our toolset for problem-solving particularly when dealing with logarithmic equations.
For example, applying this to our problem, we started with \( \log x^2 \). Through the property of logarithms, we transformed it to \( 2 \log x \). This transformation simplifies the equation and makes it easier to solve the problem step by step.
Using these properties allows us to work with simplified expressions. When the exponents in the logarithmic function appear, rely on this property to move them outside the "log" for easier manipulation. These properties expand our toolset for problem-solving particularly when dealing with logarithmic equations.
Exponential Form
Transitioning a logarithmic equation into exponential form is one of the most impactful steps in solving equations like \( \log x = 1 \).
The equation \( \log x = 1 \) is initially in logarithmic form. To convert it into an exponential equation, remember that \( \log_b a = c \) implies \( a = b^c \) in exponential form.
Since our equation is \( \log_{10} x = 1 \), we express it exponentially as \( x = 10^1 \). The conversion uncovers the value of \( x \) directly. Therefore, in our exercise, \( x = 10 \).
Understanding how to shift between logarithmic and exponential forms is crucial. It permits us to access another form of the same equation that might be simpler to interpret or solve. Utilize this method anytime you need to "undo" logarithms effectively.
The equation \( \log x = 1 \) is initially in logarithmic form. To convert it into an exponential equation, remember that \( \log_b a = c \) implies \( a = b^c \) in exponential form.
Since our equation is \( \log_{10} x = 1 \), we express it exponentially as \( x = 10^1 \). The conversion uncovers the value of \( x \) directly. Therefore, in our exercise, \( x = 10 \).
Understanding how to shift between logarithmic and exponential forms is crucial. It permits us to access another form of the same equation that might be simpler to interpret or solve. Utilize this method anytime you need to "undo" logarithms effectively.
Verification of Solutions
Verifying solutions is a vital step in mathematics to ensure accuracy. This process confirms that the obtained solution satisfies the original problem.
For our exercise, we identified \( x = 10 \) as the potential solution, and verification requires substituting this back into the original equation \( \log x^2 = 2 \).
Substituting, we get \( \log 10^2 \) which simplifies to \( \log 100 \). Knowing \( \log 100 = 2 \), since \( 10^2 = 100 \), it confirms that our calculations are correct.
Verification not only reassures that the solution is correct but also demonstrates the importance of back-substitution. It is a good habit to develop, especially for ensuring reliability in mathematical problem-solving, leaving no room for errors.
For our exercise, we identified \( x = 10 \) as the potential solution, and verification requires substituting this back into the original equation \( \log x^2 = 2 \).
Substituting, we get \( \log 10^2 \) which simplifies to \( \log 100 \). Knowing \( \log 100 = 2 \), since \( 10^2 = 100 \), it confirms that our calculations are correct.
Verification not only reassures that the solution is correct but also demonstrates the importance of back-substitution. It is a good habit to develop, especially for ensuring reliability in mathematical problem-solving, leaving no room for errors.
Other exercises in this chapter
Problem 74
Write logarithm as the sum and/or difference of logarithms of a single quantity. Then simplify, if possible. \(\ln \sqrt[3]{e^{2} x}\)
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assume that there are no deposits or withdrawals. Continuous Compound Interest. An initial investment of \(\$ 2,000\) earns \(8 \%\) interest, compounded contin
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Let \(f(x)=x^{2}-1\) and \(g(x)=x^{2}-4 .\) Find each function and give its domain. $$ g / f $$
View solution Problem 75
Write logarithmic expression as one logarithm. \(\log _{2}(x+1)+9 \log _{2} x\)
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