Problem 4
Question
Write each as an exponential equation. See Example 1. $$ \log _{5} \frac{1}{25}=-2 $$
Step-by-Step Solution
Verified Answer
The exponential equation is \(5^{-2} = \frac{1}{25}\).
1Step 1: Understand the logarithmic equation
In the equation \(\log_{5} \frac{1}{25} = -2\), it states that \(5\) is the base of the logarithm, \(\frac{1}{25}\) is the result, and \(-2\) is the exponent. Our goal is to convert this into an exponential form.
2Step 2: Recall the definition of a logarithm
The logarithmic equation \(\log_b(a) = c\) can be converted into an exponential equation \(b^c = a\). We apply this definition to the given problem.
3Step 3: Apply the definition to the given equation
Using the definition from Step 2, for \(\log_{5}\frac{1}{25} = -2\), we have the equivalent exponential form: \(5^{-2} = \frac{1}{25}\).
4Step 4: Verify the exponential equation
Calculate \(5^{-2}\) to confirm it equals \(\frac{1}{25}\). Since \(5^{-2} = \frac{1}{5^2} = \frac{1}{25}\), the conversion is verified.
Key Concepts
LogarithmsExponential FormMathematical Conversions
Logarithms
Logarithms are a mathematical concept that expresses the power to which a number, known as the base, must be raised to produce another number. In the expression \(\log_b(a) = c\), \(b\) is the base, \(a\) is the result, and \(c\) is the exponent (also known as the logarithm).
Logarithms are incredibly useful for solving equations where the unknown occurs as an exponent. They are also used to simplify calculations involving multiplication and division. Understanding how to convert logarithms to exponential form can help solve a range of mathematical problems efficiently.
When you see a logarithmic equation like \(\log_5(\frac{1}{25}) = -2\), it signifies that 5 raised to the power of -2 will result in \(\frac{1}{25}\). This conversion is key to reinterpreting logarithmic equations into a more understandable form.
Logarithms are incredibly useful for solving equations where the unknown occurs as an exponent. They are also used to simplify calculations involving multiplication and division. Understanding how to convert logarithms to exponential form can help solve a range of mathematical problems efficiently.
When you see a logarithmic equation like \(\log_5(\frac{1}{25}) = -2\), it signifies that 5 raised to the power of -2 will result in \(\frac{1}{25}\). This conversion is key to reinterpreting logarithmic equations into a more understandable form.
Exponential Form
Exponential form is a compact way of expressing repeated multiplication of a number by itself. When you've got a number like \(b^c\), it means the base \(b\) is multiplied by itself \(c\) times.
For instance, in the equation \(5^{-2} = \frac{1}{25}\), the concept of exponents helps us understand that a negative exponent like -2 means taking the reciprocal of the base raised to the positive exponent. So, \(5^{-2}\) is the reciprocal of \(5^2\), which is \(\frac{1}{5^2}\) or \(\frac{1}{25}\).
Using exponential form can simplify both the representation and manipulation of mathematical equations, making it easier to see relationships between numbers and solve logarithmic problems.
For instance, in the equation \(5^{-2} = \frac{1}{25}\), the concept of exponents helps us understand that a negative exponent like -2 means taking the reciprocal of the base raised to the positive exponent. So, \(5^{-2}\) is the reciprocal of \(5^2\), which is \(\frac{1}{5^2}\) or \(\frac{1}{25}\).
Using exponential form can simplify both the representation and manipulation of mathematical equations, making it easier to see relationships between numbers and solve logarithmic problems.
Mathematical Conversions
Mathematical conversions often involve changing the form of a mathematical expression to another form that is easier to work with. Converting between logarithmic and exponential forms is one such conversion.
This conversion hinges on the fact that logarithms are just another way to express exponents: the equation \(\log_b(a) = c\) translates to \(b^c = a\).
Understanding this conversion allows you to move seamlessly between two mathematical expressions that tell the same story in different ways. It's particularly helpful in solving equations, calculating quantities, or interpreting data in the sciences and engineering. Such proficiency in conversions enhances problem-solving skills and deepens comprehension of how different mathematical concepts are interrelated.
This conversion hinges on the fact that logarithms are just another way to express exponents: the equation \(\log_b(a) = c\) translates to \(b^c = a\).
Understanding this conversion allows you to move seamlessly between two mathematical expressions that tell the same story in different ways. It's particularly helpful in solving equations, calculating quantities, or interpreting data in the sciences and engineering. Such proficiency in conversions enhances problem-solving skills and deepens comprehension of how different mathematical concepts are interrelated.
Other exercises in this chapter
Problem 3
For the functions \(f\) and \(g,\) find \(a .(f+g)(x), b .(f-g)(x)\) \(c .(f \cdot g)(x),\) and \(d .\left(\frac{f}{g}\right)(x) .\) $$ f(x)=x^{2}+1, g(x)=5 x $
View solution Problem 3
Solve each equation. Give an exact solution and approximate the solution to four decimal places. See Example 1. $$ 3^{2 x}=3.8 $$
View solution Problem 4
Solve each equation. Give an exact solution and approximate the solution to four decimal places. See Example 1. $$ 5^{3 x}=5.6 $$
View solution Problem 4
Use a calculator to approximate each logarithm to four decimal places. See Examples I and 5. \(\log 4.86\)
View solution