Problem 92
Question
Find the exact value of the logarithm without using a calculator. If this is not possible, state the reason.$$\log _{5}\left(\frac{1}{125}\right)$$.
Step-by-Step Solution
Verified Answer
The exact value of \( \log _{5}\left(\frac{1}{125}\right) \) is \(-3\)
1Step 1 - Convert into exponential form
The formula to convert from logarithmic form to exponential form is \(b^{x}=a\), where \(b\), \(x\), and \(a\) are the base, exponent, and argument of the logarithm, respectively. Hence, converting \( \log _{5}\left(\frac{1}{125}\right) \) to its exponential form give us \(5^{x}=\frac{1}{125}\)
2Step 2 - Simplify the expression
Now, \(5^{x}=\frac{1}{125}\) can be simplified, noting that \( \frac{1}{125} = \frac{1}{5^3} = 5^{-3}\). Therefore, the previous equation simplifies to \(5^{x}=5^{-3}\)
3Step 3 - Solve for x
In the equation \(5^{x}=5^{-3}\), the bases on both sides of the equation are the same, so we can set their exponents equal: \[x = -3\]
Key Concepts
Logarithmic to Exponential FormSimplifying ExpressionsSolving Exponential Equations
Logarithmic to Exponential Form
Understanding how to convert logarithms to exponential form can be a key skill in solving problems like evaluating logs without a calculator. To make this conversion, remember the relationship between logs and exponents: a logarithm \(\log_b(a)\) answers the question, 'to what power must we raise the base \(b\) to get \(a\)?'
In the exercise \(\log _{5}\left(\frac{1}{125}\right)\), the logarithmic form reveals the base \(b = 5\) and the argument \(a = \frac{1}{125}\). To convert this to exponential form, we essentially reverse engineer this process and ask ourselves the aforementioned question. This leads us to the exponential form \(5^{x} = \frac{1}{125}\) as shared in the solution.
Using this conversion technique enables students to take a logarithmic expression and translate it into a more familiar exponential equation, preparing it for further simplification and solution, which are covered in subsequent steps.
In the exercise \(\log _{5}\left(\frac{1}{125}\right)\), the logarithmic form reveals the base \(b = 5\) and the argument \(a = \frac{1}{125}\). To convert this to exponential form, we essentially reverse engineer this process and ask ourselves the aforementioned question. This leads us to the exponential form \(5^{x} = \frac{1}{125}\) as shared in the solution.
Using this conversion technique enables students to take a logarithmic expression and translate it into a more familiar exponential equation, preparing it for further simplification and solution, which are covered in subsequent steps.
Simplifying Expressions
The art of simplifying expressions is crucial when seeking exact values in mathematics, especially without the aid of a calculator. Simplification often involves recognizing and applying the properties of numbers and operations.
For instance, when faced with \(5^{x} = \frac{1}{125}\), simplification requires recognizing that \(125\) is a power of \(5\), which we can write as \(5^3\). But, because it's in the denominator, it's equivalent to \(5^{-3}\). Thus, the expression simplifies to \(5^{x} = 5^{-3}\), which now clearly lays the groundwork for the next step: solving the exponential equation.
Being able to simplify expressions not only cleans up equations for easier handling but also significantly reduces the complexity of the problem, making the route to the solution much more straightforward and less error-prone.
For instance, when faced with \(5^{x} = \frac{1}{125}\), simplification requires recognizing that \(125\) is a power of \(5\), which we can write as \(5^3\). But, because it's in the denominator, it's equivalent to \(5^{-3}\). Thus, the expression simplifies to \(5^{x} = 5^{-3}\), which now clearly lays the groundwork for the next step: solving the exponential equation.
Being able to simplify expressions not only cleans up equations for easier handling but also significantly reduces the complexity of the problem, making the route to the solution much more straightforward and less error-prone.
Solving Exponential Equations
The process of solving exponential equations requires understanding the nature of exponents and how they can be manipulated to isolate the variable. In the given exercise, after simplifying, we end up with the equation \(5^{x} = 5^{-3}\). At this stage, solving the equation is a matter of realizing that if the bases are the same, their exponents must also be equal for the equation to be true.
This gives us the equality \(x = -3\), providing the exact value of the logarithm in question. Solving exponential equations like this one often comes down to recognizing patterns and applying basic properties of exponents. It's important for students to be comfortable with these concepts to confidently tackle and solve exponential equations across various mathematical contexts.
This gives us the equality \(x = -3\), providing the exact value of the logarithm in question. Solving exponential equations like this one often comes down to recognizing patterns and applying basic properties of exponents. It's important for students to be comfortable with these concepts to confidently tackle and solve exponential equations across various mathematical contexts.
Other exercises in this chapter
Problem 91
Think About It In Exercises \(89-92,\) place the correct symbol \(( )\) between the two numbers. $$5^{-3} \quad 3^{-5}$$
View solution Problem 92
Use a graphing utility to graph the function and approximate its zero accurate to three decimal places. $$h(t)=e^{-0.125 t}-8$$
View solution Problem 92
Think About It In Exercises \(89-92,\) place the correct symbol \(( )\) between the two numbers. $$4^{1 / 2} \quad\left(\frac{1}{2}\right)^{4}$$
View solution Problem 93
Solve the logarithmic equation algebraically. Round the result to three decimal places. Verify your answer(s) using a graphing utility. $$\ln x=-3$$
View solution