Problem 42
Question
Use the laws of logarithms to solve the equation. $$\log _{3}(x+1)+\log _{3}(2 x-3)=1$$
Step-by-Step Solution
Verified Answer
The two possible solutions for x are: \( x = 2 \) and \( x = -\frac{3}{2} \).
1Step 1: Combine logarithms using product rule
We will combine both logarithmic expressions into one using the product rule which states:
\( \log_{b}(a) + \log_{b}(c) = \log_{b}(ac) \)
So, the equation will become:
\( \log_{3}((x+1)(2x-3)) = 1 \)
2Step 2: Apply the power rule to eliminate the logarithm
Now, we will use the power rule to eliminate the logarithm. This rule states that:
\( \log_{b}(a) = c \) can be written as \( a = b^c \)
Accordingly, our equation will become:
\( (x+1)(2x-3) = 3^1 \)
3Step 3: Expand the equation and subtract the constant term
Expand the equation and subtract the constant term from both sides to get a quadratic equation:
\( (x+1)(2x-3) = 3 \)
\( 2x^2 - 3x + 2x - 3 = 3 \)
\( 2x^2 - x - 3 = 3 \)
Subtract 3 from both sides:
\( 2x^2 - x - 6 = 0 \)
4Step 4: Solve the quadratic equation
Now, we have a quadratic equation to solve. We will use the quadratic formula for this purpose:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
In our equation, \( a = 2, b = -1, c = -6 \). Hence,
\( x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-6)}}{2(2)} \)
\( x = \frac{1 \pm \sqrt{1 + 48}}{4} \)
\( x = \frac{1 \pm \sqrt{49}}{4} \)
5Step 5: Simplify and find the two possible values for x
Now, we simplify and find both possible values for x:
\( x = \frac{1 + 7}{4} \) or \( x = \frac{1 - 7}{4} \)
\( x = \frac{8}{4} \) or \( x = \frac{-6}{4} \)
\( x = 2 \) or \( x = -\frac{3}{2} \)
Therefore, the two possible solutions for x are: \( x = 2 \) and \( x = -\frac{3}{2} \).
Key Concepts
Laws of LogarithmsQuadratic EquationsProduct RulePower Rule
Laws of Logarithms
The laws of logarithms are fundamental rules that help simplify logarithmic expressions and solve logarithmic equations.
These laws include:
These laws include:
- Product Rule: This states that the logarithm of a product is the sum of the logarithms of the factors. Mathematically, \( \log_{b}(a) + \log_{b}(c) = \log_{b}(ac) \).
- Quotient Rule: This states that the logarithm of a quotient is the difference of the logarithms. It’s expressed as \( \log_{b}(\frac{a}{c}) = \log_{b}(a) - \log_{b}(c) \).
- Power Rule: This states that the logarithm of a number raised to a power is the power times the logarithm of the number, \( \log_{b}(a^n) = n \log_{b}(a) \).
Quadratic Equations
A quadratic equation is a second-degree polynomial equation of the form \( ax^2 + bx + c = 0 \). Solving these equations involves finding the values of \( x \) that make the equation zero.
Quadratic equations can be solved using various methods such as factoring, completing the square, or using the quadratic formula. The quadratic formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]In our problem, after using the laws of logarithms and expanding the expression, we arrived at a quadratic equation: \( 2x^2 - x - 6 = 0 \).
By applying the quadratic formula, we found the solution to be \( x = 2 \) and \( x = -\frac{3}{2} \). This showcases how logarithmic and algebraic techniques can be intertwined to solve complex equations.
Quadratic equations can be solved using various methods such as factoring, completing the square, or using the quadratic formula. The quadratic formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]In our problem, after using the laws of logarithms and expanding the expression, we arrived at a quadratic equation: \( 2x^2 - x - 6 = 0 \).
By applying the quadratic formula, we found the solution to be \( x = 2 \) and \( x = -\frac{3}{2} \). This showcases how logarithmic and algebraic techniques can be intertwined to solve complex equations.
Product Rule
The product rule in the context of logarithms is an elegant way to simplify expressions by combining two logarithmic terms into one.
This is incredibly helpful when solving logarithmic equations as it reduces the complexity of the expression and facilitates further manipulation.
For example, given the equation \( \log_{b}(a) + \log_{b}(c) \), the product rule allows us to rewrite it as \( \log_{b}(ac) \).
This is particularly useful when you encounter terms that can be paired together, as was done in our original exercise where we simplified \( \log_{3}(x+1) + \log_{3}(2x-3) \) into \( \log_{3}((x+1)(2x-3)) \).
Remembering this rule makes it simpler to break down and solve seemingly complex logarithmic problems.
This is incredibly helpful when solving logarithmic equations as it reduces the complexity of the expression and facilitates further manipulation.
For example, given the equation \( \log_{b}(a) + \log_{b}(c) \), the product rule allows us to rewrite it as \( \log_{b}(ac) \).
This is particularly useful when you encounter terms that can be paired together, as was done in our original exercise where we simplified \( \log_{3}(x+1) + \log_{3}(2x-3) \) into \( \log_{3}((x+1)(2x-3)) \).
Remembering this rule makes it simpler to break down and solve seemingly complex logarithmic problems.
Power Rule
The power rule of logarithms allows us to eliminate the logarithm from an equation, transforming it into a simpler algebraic expression.
The rule is stated as: \( \log_{b}(a) = c \) leads to \( a = b^c \). This means that if we know a logarithmic expression is equal to a constant, we can convert it to an exponential form.
In our exercise, after combining our original logarithmic expressions using the product rule, we were left with \( \log_{3}((x+1)(2x-3)) = 1 \).
Applying the power rule, this simplifies to \( (x+1)(2x-3) = 3^1 \). This is an essential step, as it converts our logarithmic equation into a format that can be addressed with traditional algebraic techniques. Mastery of the power rule helps streamline solving logarithmic equations by transitioning from log to exponential form efficiently.
The rule is stated as: \( \log_{b}(a) = c \) leads to \( a = b^c \). This means that if we know a logarithmic expression is equal to a constant, we can convert it to an exponential form.
In our exercise, after combining our original logarithmic expressions using the product rule, we were left with \( \log_{3}((x+1)(2x-3)) = 1 \).
Applying the power rule, this simplifies to \( (x+1)(2x-3) = 3^1 \). This is an essential step, as it converts our logarithmic equation into a format that can be addressed with traditional algebraic techniques. Mastery of the power rule helps streamline solving logarithmic equations by transitioning from log to exponential form efficiently.
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