Chapter 14: Problem 39
Solve: \(\log _{3} x+\log _{3} 2=-2\)
Short Answer
Expert verified
x = \frac{1}{18}.
Step by step solution
01
- Use Logarithm Properties
Combine the logarithms on the left-hand side using the property \(\log_b(a) + \log_b(c) = \log_b(a \cdot c)\). This gives you \(\log_3(x) + \log_3(2) = \log_3(2x)\).
02
- Rewrite the Equation
The equation now becomes \(\log_3(2x) = -2\).
03
- Convert to Exponential Form
Convert from logarithmic form to exponential form using the property \(\log_b(a) = c \implies a = b^c\). Thus, \(2x = 3^{-2}\).
04
- Simplify the Exponential Expression
Calculate \(3^{-2} = \frac{1}{3^2} = \frac{1}{9}\). So the equation now is \(2x = \frac{1}{9}\).
05
- Solve for x
Divide both sides of the equation by 2 to isolate \(x\). Thus, \(x = \frac{1}{9 \cdot 2} = \frac{1}{18}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
properties of logarithms
Understanding the properties of logarithms is essential when solving logarithmic equations. One of the key properties is the product property, which states that \(\begin{aligned} \log_b(a) + \log_b(c) = \log_b(a \cdot c)\end{aligned}\).
This property allows us to combine two logarithms into a single logarithm.
To illustrate, in the given problem \(\begin{aligned} \log_{3} x + \log_{3} 2 = -2\end{aligned}\), we use this property to combine the left side:
\(\begin{aligned} \log_{3}(x) + \log_{3}(2) \implies \log_{3}(2x)\end{aligned}\).
By converting the sum of logarithms into a single logarithm, we simplify the equation significantly.
Understanding and applying properties like these help in reducing the complexity of logarithmic equations.
Another crucial property is the quotient property, \(\begin{aligned} \log_b(a) - \log_b(c) = \log_b(a / c)\end{aligned}\).
This property is useful for equations involving subtraction of logarithms.
Lastly, the power property is also important: \(\begin{aligned} \log_b(a^c) = c \cdot \log_b(a)\end{aligned}\).
This property allows us to move an exponent in front of the logarithm.
By mastering these properties, solving logarithmic equations becomes much simpler and more intuitive.
This property allows us to combine two logarithms into a single logarithm.
To illustrate, in the given problem \(\begin{aligned} \log_{3} x + \log_{3} 2 = -2\end{aligned}\), we use this property to combine the left side:
\(\begin{aligned} \log_{3}(x) + \log_{3}(2) \implies \log_{3}(2x)\end{aligned}\).
By converting the sum of logarithms into a single logarithm, we simplify the equation significantly.
Understanding and applying properties like these help in reducing the complexity of logarithmic equations.
Another crucial property is the quotient property, \(\begin{aligned} \log_b(a) - \log_b(c) = \log_b(a / c)\end{aligned}\).
This property is useful for equations involving subtraction of logarithms.
Lastly, the power property is also important: \(\begin{aligned} \log_b(a^c) = c \cdot \log_b(a)\end{aligned}\).
This property allows us to move an exponent in front of the logarithm.
By mastering these properties, solving logarithmic equations becomes much simpler and more intuitive.
exponential form conversion
Converting logarithmic equations to exponential form is a vital step in solving them.
This method helps in isolating the variable, making it easier to find its value.
The basic property we use is:
\(\begin{aligned} \log_b(a) = c \implies a = b^c\end{aligned}\).
Let's see this in action with the problem \(\begin{aligned} \log_{3}(2x) = -2\end{aligned}\).
To convert this to exponential form, we rewrite the equation as:
\(\begin{aligned} 2x = 3^{-2}\end{aligned}\).
This transformation changes a logarithmic equation into an exponential equation, making it easier to handle.
In our case, we need to calculate \(\begin{aligned} 3^{-2} = \frac{1}{3^2} \end{aligned}\), resulting in \(\begin{aligned} 2x = \frac{1}{9} \end{aligned}\).
Converting between these forms, logarithmic to exponential and vice versa, is a foundational skill.
It provides a straightforward way to work with and solve logarithmic equations.
Once in exponential form, basic algebra can be applied to isolate and solve for the variable.
This method helps in isolating the variable, making it easier to find its value.
The basic property we use is:
\(\begin{aligned} \log_b(a) = c \implies a = b^c\end{aligned}\).
Let's see this in action with the problem \(\begin{aligned} \log_{3}(2x) = -2\end{aligned}\).
To convert this to exponential form, we rewrite the equation as:
\(\begin{aligned} 2x = 3^{-2}\end{aligned}\).
This transformation changes a logarithmic equation into an exponential equation, making it easier to handle.
In our case, we need to calculate \(\begin{aligned} 3^{-2} = \frac{1}{3^2} \end{aligned}\), resulting in \(\begin{aligned} 2x = \frac{1}{9} \end{aligned}\).
Converting between these forms, logarithmic to exponential and vice versa, is a foundational skill.
It provides a straightforward way to work with and solve logarithmic equations.
Once in exponential form, basic algebra can be applied to isolate and solve for the variable.
solving equations
Solving equations involves a sequence of logical steps to isolate the desired variable.
For the given problem, \(\begin{aligned} 2x = \frac{1}{9}\end{aligned}\):
This gives us:
\(\begin{aligned} x = \frac{1}{9 \cdot 2} = \frac{1}{18} \end{aligned}\).
Following these systematic steps ensures accuracy and helps in solving logarithmic equations.
Each step, from converting forms to applying algebraic operations, brings us closer to the solution.
Notice how every transformation is aimed at simplifying the equation to make it more manageable.
Practice these methods to improve your problem-solving skills and gain confidence in dealing with logarithmic equations.
For the given problem, \(\begin{aligned} 2x = \frac{1}{9}\end{aligned}\):
- First, simplify the expression on the exponential side.
- Calculate \(\begin{aligned} 3^{-2} = \frac{1}{9} \end{aligned}\).
- Next, isolate the variable \(\begin{aligned} x \end{aligned}\) by dividing both sides by 2.
This gives us:
\(\begin{aligned} x = \frac{1}{9 \cdot 2} = \frac{1}{18} \end{aligned}\).
Following these systematic steps ensures accuracy and helps in solving logarithmic equations.
Each step, from converting forms to applying algebraic operations, brings us closer to the solution.
Notice how every transformation is aimed at simplifying the equation to make it more manageable.
Practice these methods to improve your problem-solving skills and gain confidence in dealing with logarithmic equations.
