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Solve for \(y\) in terms of \(x\).\(\ln y=\ln (2 x+1)+\ln 1\)

Short Answer

Expert verified
The solution for \(y\) in terms of \(x\) in the given equation is \(y = 2x + 1\).

Step by step solution

01

Combine the logarithmic expressions.

Combine \(\ln (2 x +1)\) and \(\ln 1\) into a single \(\ln\) using the property \(\log(a) + log(b) = log(ab)\). The equation becomes \(\ln y = \ln ((2 x +1) \cdot 1)\), simplifying further to \(\ln y = \ln (2x + 1)\).
02

Convert equation format.

Now, eliminate the natural log \(\ln\) from both sides using the rule \( \ln a = \ln b\) if and only if \(a = b\), meaning if the natural logs of two quantities are equal, then the quantities are also equal. So the equation simplifies to \(y = 2x + 1\).
03

Verify the solution.

To verify that \(y = 2x + 1\) does indeed satisfy the original equation, substitute \(y = 2x +1\) into the original equation: \(\ln (2x +1) = \ln (2x +1) + \ln 1\). Clearly, the two sides are equal, so the solution is valid.

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Key Concepts

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

Solving Logarithmic Equations
Understanding how to solve logarithmic equations is a cornerstone in algebra and pre-calculus. A logarithmic equation contains at least one logarithm term, and the goal is to determine the unknown variable. To do this effectively, combining like terms and using the properties of logarithms can come in handy, as seen in the textbook exercise.

When faced with multiple logarithmic terms, such as \(\ln (2x + 1) + \ln 1\), our first instinct should be to apply the property that allows us to combine logarithms with the same base, which is expressed as \(\log_b(a) + \log_b(c) = \log_b(a \cdot c)\). Consistently applying these properties not only lends itself to a simpler equation but also to a clear approach to isolate the variable we are solving for.

Once combined, we can often eliminate the logarithm altogether, provided we have a logarithm on both sides with the same base, as demonstrated in Step 2 of the solution. This process transforms the logarithmic equation into a more manageable algebraic equation where traditional solving methods can be used to find the solution. It is important to remember that not all logarithmic equations are as straightforward as the example and may require additional algebraic manipulation and understanding of logarithmic functions.
Properties of Logarithms
While the previous section discussed combining logarithmic terms, it's essential to delve deeper into the properties of logarithms. These rules are not random but are closely related to the properties of exponents, since logarithms are the inverse functions of exponentials. Here are the key properties:

  • Product Rule: \(\log_b(mn) = \log_b(m) + \log_b(n)\) - used to separate or combine logs with multiplication.
  • Quotient Rule: \(\log_b(\frac{m}{n}) = \log_b(m) - \log_b(n)\) - used to separate or combine logs with division.
  • Power Rule: \(\log_b(m^n) = n \log_b(m)\) - allows us to move the exponent to the front of the log as a coefficient.
  • Change of Base Formula: \(\log_b(m) = \frac{\log_k(m)}{\log_k(b)}\) - used to convert logs of one base to another, usually to base 10 or base e for calculators.

Understanding and applying these properties enable us to manipulate and simplify logarithmic equations, making them easier to solve. It is vital for students to not just memorize these properties, but to understand them deeply, as they will reoccur throughout higher levels of mathematics.
Exponential and Logarithmic Relationships
The connection between exponential and logarithmic functions is inherent in their definitions. An exponential function like \(b^x\) involves raising a constant base \(b\) to a variable exponent \(x\), whereas a logarithmic function like \(\log_b(y)\) asks the question 'to what exponent must we raise \(b\) to obtain \(y\)?' Mathematically, these concepts are inverse operations of each other.

To illustrate, if we have an equation \(b^x = y\), we can write the equivalent logarithmic form as \(x = \log_b(y)\). Likewise, if we start with \(\log_b(y) = x\), we can express this in its exponential form as \(b^x = y\). This is the crux of how we can remove logarithms from an equation when solving, as we leverage their inverse relationship.

When confronted with logarithmic equations, it's useful to remember this relationship—especially when verifying solutions. By converting between log and exponential forms, we can better understand the equation we're working with and ensure our solutions are accurate, much like the verification step in the textbook solution provided. Internalizing this relationship is a powerful tool in a student's mathematical arsenal.

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Most popular questions from this chapter

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(-14+3 e^{x}=11\)

Domestic Demand The domestic demands \(D\) (in thousands of barrels) for refined oil products in the United States from 1995 to 2005 are shown in the table. (Source: U.S. Energy Information Administration)$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Demand } \\ \hline 1995 & 6,469,625 \\ \hline 1996 & 6,701,094 \\ \hline 1997 & 6,796,300 \\ \hline 1998 & 6,904,705 \\ \hline 1999 & 7,124,435 \\ \hline 2000 & 7,210,566 \\ \hline \end{array} $$$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Demand } \\ \hline 2001 & 7,171,885 \\ \hline 2002 & 7,212,765 \\ \hline 2003 & 7,312,410 \\ \hline 2004 & 7,587,546 \\ \hline 2005 & 7,539,440 \\ \hline \end{array} $$(a) Use a spreadsheet software program to create a scatter plot of the data. Let \(t\) represent the year, with \(t=5\) corresponding to 1995 . (b) Use the regression feature of a spreadsheet software program to find an exponential model for the data. Use the Inverse Property \(b=e^{\ln b}\) to rewrite the model as an exponential model in base \(e\). (c) Use the regression feature of a spreadsheet software program to find a logarithmic model \((y=a+b \ln x)\) for the data. (d) Use a spreadsheet software program to graph the exponential model in base \(e\) and the logarithmic model with the scatter plot. (e) Use both models to predict domestic demands in 2008 , 2009, and \(2010 .\) Do both models give reasonable predictions? Explain.

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(1000 e^{-4 x}=75\)

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Solve the exponential equation algebraically. Approximate the result to three decimal places.\(8\left(4^{6-2 x}\right)+13=41\)

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