Problem 69
Question
Solve each logarithmic equation in Exercises \(49-92 .\) 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. $$ \log _{3}(x+6)+\log _{3}(x+4)=1 $$
Step-by-Step Solution
Verified Answer
The solution to the equation \(\log _{3}(x+6)+\log _{3}(x+4)=1\) is \(x = -3\).
1Step 1: Combine logarithmic expressions
The sum of logarithms with the same base (\(3\) in this case) can be rewritten as a single logarithm by using the logarithmic rule \(\log_b(m) + \log_b(n) = \log_b(m * n)\). Using this rule, our equation becomes \(\log_3((x+6) * (x+4)) = 1\).
2Step 2: Conversion to exponential form
The equation can now be converted to an exponential form. Logarithmic form and exponential form are two ways to represent the same equation and we can switch between the two. The transformation is based on the rule \(\log_b(a) = c\) is equivalent to \(b^c = a\). Thus the equation is transformed to \(3^1 = (x+6)*(x+4)\).
3Step 3: Solving for \(x\)
This gives the quadratic equation \(x^2 + 10x + 24 - 3 = 0\) or \(x^2 + 10x + 21 = 0\). This quadratic equation can be factored into \((x + 7)(x + 3) = 0\). The solutions to this equation are \(x = -7\) and \(x = -3\).
4Step 4: Checking the domain
Finally, the found values of \(x\) need to be checked to see if they are in the domain of the original logarithmic expressions. Since \(\log_b(a)\) is not defined for \(a \leq 0\), we can reject \(x = -7\) since this is not in the domain of \(x+6\) and \(x+4\). The final valid solution is \(x = -3\).
Key Concepts
Logarithmic ExpressionsExponential FormQuadratic EquationDomain of Logarithmic Function
Logarithmic Expressions
Understanding logarithmic expressions is essential when trying to solve equations involving logs. A logarithm, represented as \( \log_b(a) \), refers to the power to which a base, \( b \), must be raised to obtain a number, \( a \). The logarithmic expression essentially asks the question, 'To what exponent \( x \), must we raise \( b \), to get \( a \)?'
For example, to combine logarithmic expressions, we can apply certain logarithmic properties. The textbook exercise given demonstrates a situation where we have two logs with the same base being added: \( \log_{3}(x+6) + \log_{3}(x+4) \). Using the logarithmic property where the sum of two logs (with the same base) is equal to the log of the product of their arguments, we can rewrite it as: \( \log_{3}((x+6)(x+4)) \). This simplification is the first crucial step in solving the given logarithmic equation.
For example, to combine logarithmic expressions, we can apply certain logarithmic properties. The textbook exercise given demonstrates a situation where we have two logs with the same base being added: \( \log_{3}(x+6) + \log_{3}(x+4) \). Using the logarithmic property where the sum of two logs (with the same base) is equal to the log of the product of their arguments, we can rewrite it as: \( \log_{3}((x+6)(x+4)) \). This simplification is the first crucial step in solving the given logarithmic equation.
Exponential Form
Converting logarithmic expressions to exponential form can sometimes clarify the solution process. In exponential form, the bases and exponents are made explicit: if we have a logarithm \( \log_b(a) = c \), it can be rewritten as \( b^c = a \). This relationship shows the direct connection between logs and exponentials; both are different ways to express the same idea.
In the context of the exercise, once the logarithmic expressions have been combined, the next step is to use this conversion to create an equation that may be solved more straightforwardly. We take \( \log_{3}((x+6)(x+4)) = 1 \) and convert it to \( 3^1 = (x+6)(x+4)\), setting up a quadratic equation that will help us find the value of \( x \).
In the context of the exercise, once the logarithmic expressions have been combined, the next step is to use this conversion to create an equation that may be solved more straightforwardly. We take \( \log_{3}((x+6)(x+4)) = 1 \) and convert it to \( 3^1 = (x+6)(x+4)\), setting up a quadratic equation that will help us find the value of \( x \).
Quadratic Equation
A quadratic equation is in the form of \( ax^2 + bx + c = 0 \) wherein \( a \) , \( b \) , and \( c \) are constants, and \( x \) is the unknown variable. Quadratic equations often arise after applying algebraic manipulations to various problems, such as the one we're reviewing. To solve a quadratic equation, one can factorize it, complete the square, or use the quadratic formula.
The exercise leads us to the quadratic equation \( x^2 + 10x + 21 = 0 \) after converting the logarithmic equation to exponential form and expanding the right side. This equation is then factored into \( (x + 7)(x + 3) = 0 \) to find the potential solutions for \( x \). Each factor represents a possible solution to the equation, which we can test against the original equation's domain.
The exercise leads us to the quadratic equation \( x^2 + 10x + 21 = 0 \) after converting the logarithmic equation to exponential form and expanding the right side. This equation is then factored into \( (x + 7)(x + 3) = 0 \) to find the potential solutions for \( x \). Each factor represents a possible solution to the equation, which we can test against the original equation's domain.
Domain of Logarithmic Function
The domain of a logarithmic function consists of all the possible values of \( x \) that make the function defined and real. For the function \( \log_b(x) \), the domain is \( x > 0 \) because a logarithm of a negative number or zero is undefined in the real number system.
Upon solving the quadratic equation in the exercise, we find two potential solutions: \( x = -7 \) and \( x = -3 \). However, not both will fit within the domain of the original logarithmic expressions. Since both \( x+6 \) and \( x+4 \) must be positive for the logs to be defined, \( x = -7 \) is extraneous and must be rejected. The solution \( x = -3 \) is valid as it satisfies \( x+6 > 0 \) and \( x+4 > 0 \) and thus lies within the function's domain, making \( x = -3 \) the correct solution to the logarithmic equation.
Upon solving the quadratic equation in the exercise, we find two potential solutions: \( x = -7 \) and \( x = -3 \). However, not both will fit within the domain of the original logarithmic expressions. Since both \( x+6 \) and \( x+4 \) must be positive for the logs to be defined, \( x = -7 \) is extraneous and must be rejected. The solution \( x = -3 \) is valid as it satisfies \( x+6 > 0 \) and \( x+4 > 0 \) and thus lies within the function's domain, making \( x = -3 \) the correct solution to the logarithmic equation.
Other exercises in this chapter
Problem 68
The formula \(S-C(1+r)^{t}\) models inflation, where \(C\) - the value today, \(r-\) the annual inflation rate, and \(S\) - the inflated value \(t\) years from
View solution Problem 69
The figure shows the graph of \(f(x)=\ln x\). Use transformations of this graph to graph each function. Graph and give equations of the asymptotes. Use the grap
View solution Problem 69
In Exercises \(41-70,\) use properties of logarithms to condense each logarithmic expression. Write the expression as a single logarithm whose coefficient is \(
View solution Problem 70
Solve each logarithmic equation in Exercises \(49-92 .\) Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions.
View solution