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Chapter 13: Problem 36

Determine the value of the unknown. $$\log _{7} y=-2$$

Short Answer

Expert verified
The value of \( y \) is \( \frac{1}{49} \).

Step by step solution

01

Understand the Logarithmic Equation

The equation provided is \( \log_{7} y = -2 \). In any logarithmic equation of the form \( \log_{b} x = c \), \( b \) is the base, \( x \) is the result of the base raised to the power \( c \). Here, \( y \) is the unknown, and it is in the base of 7.
02

Convert to Exponential Form

To find the value of \( y \), convert the logarithmic equation into its equivalent exponential form. This conversion follows the logic that if \( \log_{b} x = c \), then \( x = b^{c} \). Therefore, \( y = 7^{-2} \).
03

Calculate the Exponential Value

Now, calculate \( 7^{-2} \). A negative exponent indicates a reciprocal, so \( 7^{-2} = \frac{1}{7^{2}} \). Calculate \( 7^2 \), which is \( 49 \), thus \( 7^{-2} = \frac{1}{49} \).
04

Confirm the Solution by Substitution

To verify the solution, substitute \( y = \frac{1}{49} \) back into the original equation: \( \log_{7} \left( \frac{1}{49} \right) \). Since \( 49 = 7^2 \), \( \frac{1}{49} = 7^{-2} \), which is equivalent to \( -2 \). The original equation \( \log_{7} y = -2 \) is satisfied.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Form
The exponential form is a way to express numbers using a base and an exponent. It is crucial in solving logarithmic equations. Let's break this down into simple steps. When you see an equation like \( \log_b x = c \), it can be translated into exponential form by recognizing that \( x \) is the result of raising the base \( b \) to the exponent \( c \).
  • Base: The number that is repeatedly multiplied.
  • Exponent: Indicates how many times the base is multiplied by itself.
  • Translation: \( \log_{b} x = c \) translates to \( x = b^{c} \).
Understanding this conversion is key to finding unknown values in logarithmic equations.
For example, if \( \log_{7} y = -2 \), then by applying the exponential form, it translates to \( y = 7^{-2} \). This approach helps eliminate the logarithm and allows us to solve for \( y \) directly.
Negative Exponents
Negative exponents can be a bit tricky, but don't worry! Breaking them down can make them much easier to understand. A negative exponent indicates that the base should be taken as the reciprocal and then raised to the positive value of that exponent.
  • Reciprocal: Take the inverse of the base, making \( a^{-n} \) become \( \frac{1}{a^n} \).
  • Example: \( 7^{-2} = \frac{1}{7^2} \).
So when you encounter \( 7^{-2} \), it means you should first find the reciprocal of 7, making it \( \frac{1}{7} \), and then square the result, which translates to \( \frac{1}{49} \).
This principle helps simplify problems involving powers, making calculations easier and more intuitive.
Logarithmic Functions
Logarithmic functions are the inverse of exponential functions, providing a different way to look at the relationships between numbers. Understanding logarithmic functions allows you to solve equations where the unknown is in the exponent, which isn't straightforward to calculate otherwise.
  • Understanding Logs: If \( \log_{b} x = c \), then \( x = b^{c} \). This shows how logarithms and exponents intertwine.
  • Base and Argument: The base \( b \) is the number being raised to a power, and the argument \( x \) is the result of that base raised to the power \( c \).
  • Inversion: Logarithms essentially "undo" the exponentiation of the base and provide the exponent as the result.
These functions are extremely useful in various mathematics fields, allowing complex exponential relationships to be simplified and solved efficiently. In our case, solving \( \log_{7} y = -2 \) by converting it to an exponential form gives a clear path to finding the value of \( y \).

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Most popular questions from this chapter

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