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Solve for \(x\) in terms of \(y\) (a) \(\log _{10} x-y=\log _{10}(3 x-1);\) (b) \(\log _{10}(x-y)=\log _{10}(3 x-1).\)

Short Answer

Expert verified
(a) \(x = \frac{10^y}{1 - 3 \times 10^y}\) if \(y \neq \log_{10}\frac{1}{3}\); (b) \(x = \frac{1 - y}{2}\).

Step by step solution

01

Understand the Problem for Part (a)

We need to express \( x \) in terms of \( y \) using the given equation: \( \log_{10} x - y = \log_{10} (3x - 1) \). This involves solving an equation involving logarithms.
02

Simplify the Logarithmic Equation for Part (a)

First, move the \( y \) term to the right side, resulting in \( \log_{10} x = y + \log_{10} (3x - 1) \).
03

Use Logarithmic Properties for Part (a)

Apply the property \( a + b = \log_{b}(c) + \log_{b}(d) \Rightarrow \log_{b}(cd) \). Thus, \( \log_{10} x = \log_{10} [(3x - 1) \cdot 10^y] \).
04

Convert the Equation for Part (a) into an Exponential Form

Since the bases are the same, equate the arguments: \( x = (3x - 1) \cdot 10^y \).
05

Isolate \( x \) for Part (a)

Rearrange the equation from the previous step: \( x - 3x \cdot 10^y = -10^y \). Simplify it to \( x(1 - 3 \cdot 10^y) = 10^y \). Finally, \( x = \frac{10^y}{1 - 3 \cdot 10^y} \) when \( y = \frac{1}{3} \).
06

Verify the Constraints for Part (a)

Ensure the denominator is not zero. The value is valid if \( 1 - 3 \cdot 10^y eq 0 \), or \( y eq \log_{10}\frac{1}{3} \).
07

Address Part (b) Problem Statement

For part (b), the problem states \( \log_{10} (x-y) = \log_{10} (3x - 1) \). We solve for \( x \) using this equation.
08

Remove Logarithms by Equating Arguments for Part (b)

Because the logarithms are equal, equate their arguments: \( x - y = 3x - 1 \).
09

Solve for \( x \) in Terms of \( y \) for Part (b)

Rearrange to solve for \( x \): \( x - 3x = -1 + y \) gives \( -2x = -1 + y \). Simplifying, \( x = \frac{1 - y}{2} \).

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Key Concepts

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

Solving Equations
Solving equations, especially those involving logarithms, requires a systematic approach. The main goal is to express the variable of interest in terms of other given variables. This process often involves rearranging and simplifying the equation to isolate the desired variable. For example, in the equation \( \log_{10} x - y = \log_{10} (3x - 1) \), our target is to solve for \( x \) in terms of \( y \).
The key skill here is to understand how to manipulate the equation using algebraic techniques to simplify and rearrange terms. Once you have a clear expression for the variable you're solving for, you can check the constraints to ensure the solution is valid. This involves verifying that the values do not lead to undefined mathematical expressions, such as division by zero.
Logarithm Properties
Logarithms have specific properties that make them highly useful in simplifying and solving equations. Knowing these properties well can significantly ease the process of solving logarithmic equations. One important property is the ability to combine logarithms using addition:
  • If \( a + b = \log_{b}(c) + \log_{b}(d) \), then this can be rewritten as \( \log_{b}(cd) \).
  • This property helps to condense complex logarithmic expressions into simpler ones.
In our exercise, we used this property to combine the terms into a single logarithmic expression, which then allowed us to simplify and ultimately solve the equation. Understanding and applying logarithmic properties skillfully is crucial for efficient problem-solving in logarithmic equations.
Exponential Form Conversion
Converting logarithmic equations into exponential form is a powerful step in solving them. This conversion uses the principle that if \( \log_{b}(x) = y \), then \( x = b^y \). This is because a logarithm essentially represents an exponent.
  • In our exercise for part (a), the logarithmic equation \( \log_{10} x = y + \log_{10} (3x - 1) \) was converted into an exponential form \( x = (3x - 1) \cdot 10^y \).
  • Once in exponential form, the equation can be solved algebraically by rearranging and isolating \( x \).
This approach can greatly simplify complex logarithmic equations, making them more straightforward to solve. It's important to become comfortable with moving between logarithmic and exponential forms to tackle a variety of logarithmic problems effectively.

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