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91Ó°ÊÓ

Solve each equation. Give the exact answer. $$\log _{3}(x-1)=2$$

Short Answer

Expert verified
The solution is \(x = 10\).

Step by step solution

01

Understanding the Logarithmic Equation

We start with the equation \(\log _{3}(x-1)=2.\) This is a logarithmic equation where the base of the logarithm is 3, and the argument is \(x - 1\). Our goal is to solve for \(x\) by eliminating the logarithm.
02

Rewriting in Exponential Form

To eliminate the logarithm, we convert the logarithmic equation into its equivalent exponential form. Recall that \(\log_b(a) = c\) can be rewritten as \(b^c = a\). Applying this to our equation: \(3^2 = x - 1\).
03

Calculating the Exponentiation

Calculate \(3^2\) to simplify the equation further. We know that \(3^2 = 9\), so the equation becomes \(9 = x - 1\).
04

Solving for x

To find \(x\), add 1 to both sides of the equation to isolate \(x\). This gives us \(9 + 1 = x\), resulting in \(x = 10\).

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Key Concepts

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

Exponential Form
Converting a logarithmic equation to exponential form is a core step needed to solve many types of logarithmic equations. A logarithmic equation such as \( \log_b(a) = c \) can be transformed into its exponential counterpart by reinterpreting the relationship among the numbers involved. This conversion utilizes the definition of logarithms, which states that if \( \log_b(a) = c \), then the base \( b \) raised to the power \( c \) equals \( a \). This means \( b^c = a \).
To apply this knowledge, consider the exercise problem \( \log_3(x-1) = 2 \). By converting to exponential form, we rewrite it as \( 3^2 = x - 1 \). Now, the equation is easier to handle as a simpler equation involving exponents. Transforming logarithmic expressions this way allows for the effective application of algebraic techniques to isolate and solve for unknown variables. Understanding this linkage between logarithms and exponentiation is crucial for solving logarithmic equations efficiently.
Solving Equations
With the exponential form in place, our next task is solving the resulting equation to find the variable's value. Solving equations generally involves performing operations to isolate the variable.
For the exponential equation we obtained, \( 3^2 = x - 1 \), the procedure is straightforward:
  • First, calculate the exponential part. Here \( 3^2 \) equals 9.
  • Next, substitute what you found back into the equation giving \( 9 = x - 1 \).
Now proceed to isolate \( x \) by performing algebraic operations:
  • Add 1 to both sides of the equation to isolate \( x \): \( 9 + 1 = x \).
  • The solution, after adding, becomes \( x = 10 \).
Thus, by systematically applying arithmetic maneuvers, the solution to the equation emerges. The phase of solving equations requires focus on maintaining the equality while manipulating terms to solve for the variable efficiently.
Mathematical Transformation
The art of mathematical transformation is the foundation of converting and simplifying equations. This concept involves changing the form of an equation to reveal solutions more clearly.
In the context of logarithmic equations, transformations allow us to switch between logarithmic and exponential forms, as demonstrated in the problem-solving process.
  • Transforming \( \log_3(x-1) = 2 \) into \( 3^2 = x - 1 \) leverages the power of exponents to demystify the logarithmic expression.
  • These transformations simplify complex expressions and uncover simpler paths to arriving at solutions.
Mathematical transformations reinforce the notion that different algebraic expressions can describe the same relationship. Learning how to effectively use these transformations can make complex problems approachable and more easily navigated. This skill not only aids in solving equations but also contributes to a deeper understanding of mathematical concepts beyond surface arithmetic.

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Most popular questions from this chapter

Use a graph with the given viewing window to decide which functions are one- to-one. If a function is one-to-one, give the equation of its inverse function. Check your work by graphing the inverse function on the same coordinate axes. $$f(x)=\frac{x-5}{x+3} ;[-6.6,6.6] \text { by }[-4.1,4.1]$$

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Although a function may not be one-to-one when defined over its "natural" domain, it may be possible to restrict the domain in such a way that it is one-to-one and the range of the function is unchanged. For example, if we nestrict the domain of the function \(f(x)=x^{2}\) (which is not one-to-one over \((-\infty, \infty)\) to \([0, \infty)\), we obtain a one-to-one function whose range is still \([0, \infty)\) See the figure to the right. Notice that we could choose to restrict the domain of \(f(x)=x^{2}\) to \((-\infty, 0]\) and also obtain the graph of a one-to-one function, except that it would be the left half of the parabola. For each function in Exercises \(117-122\), restrict the domain so that the function is one-to-one and the range is not changed. You may wish to use a graph to help decide. Answers may vary. (GRAPHS CANNOT COPY) $$f(x)=x^{4}+x^{2}-6$$

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