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Solve for \(y\) in terms of \(x\).\(\log _{10}(y-4)+\log _{10} x=3 \log _{10} x\)

Short Answer

Expert verified
The solution to the problem is \(y = x^2 + 4\).

Step by step solution

01

Combine Logarithm Rules

First, we use the logarithm properties to simplify the equation. According to the log properties, if we have \(\log_b(m) + \log_b(n)\), we can combine these two logs into \(\log_b(mn)\). So, the equation becomes: \(\log _{10}[(y-4)x] = 3\log _{10} x\)
02

Apply Power Rule to Logarithm

Then, apply the power rule of logarithms, which states that \(a \log_b(m) = \log_b(m^a)\). So, the equation becomes: \(\log _{10}[(y-4)x] = \log _{10}(x^3)\)
03

Cancel out Logarithm

Logarithms with the same base cancel out, so if we have \(\log _{10}(a) = \log _{10}(b)\) it means that \(a = b\). Thus, we have '((y-4)x = x^3)'.
04

Solve for y

Finally, solve for \(y\), we have \(y = \frac{x^3}{x} + 4 = x^2 + 4\).

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Key Concepts

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

Logarithm Properties
Understanding the properties of logarithms can make solving logarithmic equations much easier. One of the essential properties used in solving the given problem is the property of the sum of logarithms. When you have \(\log_b(m) + \log_b(n)\), you can combine these into a single logarithm as \(\log_b(mn)\). This is called the "product rule" for logarithms. It’s useful for simplifying expressions that add two or more logarithms with the same base.

In the original problem, we used this property to combine \(\log_{10}(y-4) + \log_{10}x\) into \(\log_{10}[(y-4)x]\). Recognizing opportunities to use logarithmic properties can significantly streamline the process of manipulating and simplifying equations.

Another vital property is when you encounter a logarithmic equation like \(\log_{10}(a) = \log_{10}(b)\). This property states that if two logs with the same base are equal, then the values inside them are equal as well, i.e., \(a = b\). This allows us to drop the logarithms entirely and solve a simpler algebraic equation instead.
Power Rule
The power rule of logarithms is another crucial tool that aids in solving logarithmic expressions. According to this rule, multiplying a logarithm by a real number can be rewritten as raising the argument of the logarithm to the power of that real number.

In mathematical terms, the power rule can be expressed as \(a \log_b(m) = \log_b(m^a)\).

This rule is particularly handy when you encounter an equation like \(3\log_{10}x\). You can rewrite this as \(\log_{10}(x^3)\).

The power rule simplifies equations, making them easier to solve by allowing you to express exponential growth within the logarithm’s argument. In our example, applying this rule transformed the right side of the equation from \(3\log_{10} x\) to \(\log_{10}(x^3)\), making it easier to equate with the left side.
Solving Equations
Solving logarithmic equations involves several strategic steps. After simplifying the given logarithmic expressions using properties like the product rule and power rule, the next phase is to equate the simplified expressions and solve the algebraic equation.

In our example, after simplification, we were left with the equation \(\log_{10}[(y-4)x] = \log_{10}(x^3)\). At this point, because both sides have the same logarithmic base, \(\log_{10}\), we can directly equate their arguments: \((y-4)x = x^3\).

Now, you're left with a simpler equation to solve: \(y-4 = x^2\), after dividing both sides by \(x\), assuming \(x eq 0\). Finally, adding 4 to both sides gives the solution for \(y\), which is \(y = x^2 + 4\).

The steps involve reducing the logarithmic equation to a basic algebraic form and then using standard algebraic methods to isolate and solve for the variable. Always check the domain of the solution to ensure that it is valid within the context of the original logarithmic equation.

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Most popular questions from this chapter

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.

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