Problem 86
Question
Solve each logarithmic equation. Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ \log (x-2)+\log 5=\log 100 $$
Step-by-Step Solution
Verified Answer
The solution to the equation is \(x= 22\).
1Step 1: Simplify the equation using properties of logarithms
Using the product rule of logarithms, \(\log_a mn = \log_a m + \log_a n\), the equation could be rewritten as \(\log [(x-2)5] = \log 100\). Thus, the equation becomes \(\log (5x -10) = \log 100\).
2Step 2: Solve for x
If \(\log_a m = \log_a n\), then \(m =n\). This equality allows us to equate the arguments of the logarithms in the above expression, giving \(5x -10 = 100\). Solving this simple equation yields \(x = \(\frac{100+10}{5} =22\).
3Step 3: Verify the solution
Replace \(x\) with 22 in the original equation to verify if it satisfies the equation: \(\log (22-2) + \log 5 = \log 100\), which simplifies to \(\log 20+ \log 5 = \log 100\)\. As \(\log 100 = \log 20+\log 5\), the equation is correct, and so \(x=22\) is the solution.
Key Concepts
Properties of LogarithmsLogarithmic Equation VerificationDomain of Logarithmic Functions
Properties of Logarithms
Understanding the properties of logarithms is essential to solve logarithmic equations. Logarithms, the inverse operations to exponents, have unique properties that allow us to manipulate and simplify expressions.
One of these properties is the Product Rule, which states that \( \log_a (mn) = \log_a m + \log_a n \), where \( a \) is the base of the logarithm, and \( m \) and \( n \) are positive real numbers. This property is used to combine two logs with the same base into one, by converting their addition into the logarithm of the product of their arguments.
Similarly, the Quotient Rule expresses \( \log_a (\frac{m}{n}) = \log_a m - \log_a n \), useful for dividing two logs. Another property is the Power Rule, which states that \( \log_a (m^n) = n\log_a m \) and allows us to move the exponent of the argument to the front of the log. These rules are not only helpful in simplifying expressions but also crucial when solving logarithmic equations, as seen in the provided exercise.
Additional properties include Change of Base Formula, allowing conversion between logarithm bases, and the understanding that \( \log_a 1 = 0 \) and \( \log_a a = 1 \) for any positive \( a \) different from 1. Grasping these rules facilitates the manipulation of logarithmic statements in the process of finding solutions to complex equations.
One of these properties is the Product Rule, which states that \( \log_a (mn) = \log_a m + \log_a n \), where \( a \) is the base of the logarithm, and \( m \) and \( n \) are positive real numbers. This property is used to combine two logs with the same base into one, by converting their addition into the logarithm of the product of their arguments.
Similarly, the Quotient Rule expresses \( \log_a (\frac{m}{n}) = \log_a m - \log_a n \), useful for dividing two logs. Another property is the Power Rule, which states that \( \log_a (m^n) = n\log_a m \) and allows us to move the exponent of the argument to the front of the log. These rules are not only helpful in simplifying expressions but also crucial when solving logarithmic equations, as seen in the provided exercise.
Additional properties include Change of Base Formula, allowing conversion between logarithm bases, and the understanding that \( \log_a 1 = 0 \) and \( \log_a a = 1 \) for any positive \( a \) different from 1. Grasping these rules facilitates the manipulation of logarithmic statements in the process of finding solutions to complex equations.
Logarithmic Equation Verification
Once we solve a logarithmic equation, it's important to verify if the obtained solution is correct. This process includes ensuring the solution fits the original equation and that it is within the domain of the logarithmic functions involved.
To verify a logarithmic equation, you must replace the variable with the solution you've found, then simplify the equation. If the left and the right side of the equation are equal after the substitution, your solution is valid. In the exercise example, after solving the equation, the solution \( x = 22 \) was obtained. Substituting \( x \) back into the original equation gives us \( \log (22-2) + \log 5 = \log 100 \), which simplifies to a true statement.
However, an important note is that we must also ensure the solution does not result in taking the log of a negative number or zero, as both are undefined. In the example, the domain for \( x \) must be such that \( x-2 > 0 \) since the logarithmic function is only defined for positive arguments. Therefore, we must reject any solution that does not meet this domain criterion, ensuring all solutions are within the realm of mathematical validity.
To verify a logarithmic equation, you must replace the variable with the solution you've found, then simplify the equation. If the left and the right side of the equation are equal after the substitution, your solution is valid. In the exercise example, after solving the equation, the solution \( x = 22 \) was obtained. Substituting \( x \) back into the original equation gives us \( \log (22-2) + \log 5 = \log 100 \), which simplifies to a true statement.
However, an important note is that we must also ensure the solution does not result in taking the log of a negative number or zero, as both are undefined. In the example, the domain for \( x \) must be such that \( x-2 > 0 \) since the logarithmic function is only defined for positive arguments. Therefore, we must reject any solution that does not meet this domain criterion, ensuring all solutions are within the realm of mathematical validity.
Domain of Logarithmic Functions
The domain of logarithmic functions plays a critical role in identifying valid solutions for logarithmic equations. A logarithmic function \( \log_a x \) is defined only for positive values of \( x \) because there is no real number that \( a \) (where \( a > 0 \) and \( a eq 1 \) ) can be raised to produce a negative number or zero.
For an equation like \( \log (x-2) + \log 5 = \log 100 \), it's implied that both \( x-2 \) and 5 must be within the domain of the logarithmic function—positive numbers. This restriction is crucial during the solving process because it can invalidate solutions that otherwise algebraically satisfy the equation. For instance, if a solution to a given logarithmic equation results in a negative argument within a logarithm, that solution cannot be valid.
In the example given, the domain is all real numbers greater than 2, as \( x-2 \) must be positive. A thorough understanding of the domain ensures that solutions obtained don't only mathematically solve the equation but are also legitimate when placed within the context of logarithmic functions.
For an equation like \( \log (x-2) + \log 5 = \log 100 \), it's implied that both \( x-2 \) and 5 must be within the domain of the logarithmic function—positive numbers. This restriction is crucial during the solving process because it can invalidate solutions that otherwise algebraically satisfy the equation. For instance, if a solution to a given logarithmic equation results in a negative argument within a logarithm, that solution cannot be valid.
In the example given, the domain is all real numbers greater than 2, as \( x-2 \) must be positive. A thorough understanding of the domain ensures that solutions obtained don't only mathematically solve the equation but are also legitimate when placed within the context of logarithmic functions.
Other exercises in this chapter
Problem 86
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