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Chapter 3: Problem 58

Solve each logarithmic equation. Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$\log _{7}(x+2)=-2$$

Short Answer

Expert verified
The exact solution to the equation \(\log _{7}(x+2)=-2\) is \(x=-1.97959\)

Step by step solution

01

Conversion

Convert the logarithmic equation to an exponential equation using the basic logarithm definition. This transforms the equation \(\log _{7}(x+2)=-2\) into \(7^{-2} = x+2\).
02

Calculation

Evaluate the exponential on the left side of equation which results in \(\frac{1}{49} = x + 2\).
03

Solve for \(x\)

Isolate \(x\) on one side by subtracting \(2\) from both sides of the equation to find the value of \(x\). This gives \(x = \frac{1}{49} - 2\).
04

Simplify the Result

Simplify \(\frac{1}{49} - 2\) in the equation \(x = \frac{1}{49} - 2\) to get the exact value of \(x\). The answer is \(x=-1.97959\).
05

Check the Solution

Ensure that \(x\) is within the domain of the original logarithmic function. As \(x=-1.97959\) does not result in a negative value under the logarithm, it is indeed the solution and does not have to be rejected.

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Key Concepts

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

Domain of Logarithmic Functions
Before solving a logarithmic equation, it's crucial to determine its domain. The domain of a logarithmic function consists of all values that make the argument of the logarithm positive. In this case, the function is \(\log_{7}(x + 2)\). For the logarithm to be defined:
  • The expression \(x + 2\) must be greater than 0.
  • This leads to the inequality \(x > -2\).
Thus, the domain of the equation \(\log_{7}(x + 2) = -2\) is \(x > -2\). Any value of \(x\) that isn't greater than \(-2\) would make the logarithmic expression undefined. Knowing the domain helps us ensure that our solution is valid.
Exponential Equations
Logarithmic equations can often be solved by converting them into exponential forms. The equation \(\log_{7}(x+2)=-2\) can be rewritten, using the definition of a logarithm, as an exponential equation \(7^{-2} = x+2\).
This step leverages the property that the logarithm \(\log_b(a)=c\) translates to the exponential equation \(b^c=a\).
Recognizing this key relationship is important to navigate smoothly between logarithmic and exponential forms. It simplifies the solving process and allows us to proceed with familiar algebraic techniques.
Solving Equations
After converting the logarithmic equation to an exponential form, we proceed to solve for \(x\) using algebra. Here, we have \(7^{-2} = x + 2\). Evaluating the exponential expression gives:
  • \(7^{-2} = \frac{1}{49}\).
Now, we isolate \(x\) by subtracting \(2\) from both sides:
  • \(x = \frac{1}{49} - 2\).

This results in \(x = -1.97959\), which we further verify by substituting back into the argument of the original logarithm. Ensuring that this value is within the defined domain guarantees the solution is valid.
Decimal Approximations
Exact solutions in equations provide a precise answer, but in practical scenarios, decimal approximations give us more understandable values. When you perform the subtraction \(\frac{1}{49} - 2\), you arrive at the exact value of \(x\) first.
Then, to convert this into a decimal approximation, it's essential to be accurate.
  • Calculate \(\frac{1}{49}\), which is approximately \(0.02041\).
  • Then, subtract \(2\) to get \(-1.97959\).
For reporting purposes, you often round this number to two decimal places, thus \(x = -1.98\). Approximating helps simplify complex calculations, making them more comprehensible while still reflecting a close representation of the exact solution.

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Most popular questions from this chapter

Determine whether each statement makes sense or does not make sense, and explain your reasoning. Because the equations $$\log (3 x+1)=5 \quad \text { and } \quad \log (3 x+1)=\log 5$$ are similar, I solved them using the same method.

Use your graphing utility to graph each side of the equation in the same viewing rectangle. Then use the \(x\) -coordinate of the intersection point to find the equation's solution set. Verify this value by direct substitution into the equation. $$2^{x+1}=8$$

determine whether each statement makes sense or does not make sense, and explain your reasoning. I've noticed that exponential functions and logarithmic functions exhibit inverse, or opposite, behavior in many ways. For example, a vertical translation shifts an exponential function's horizontal asymptote and a horizontal translation shifts a logarithmic function's vertical asymptote.

Hurricanes are one of nature's most destructive forces. These low-pressure areas often have diameters of over 500 miles. The function \(f(x)=0.48 \ln (x+1)+27\) models the barometric air pressure, \(f(x),\) in inches of mercury, at a distance of \(x\) miles from the eye of a hurricane. Use this function to solve. Use an equation to answer this question: How far from the eye of a hurricane is the barometric air pressure 29 inches of mercury? Use the [TRACE] and [ZOOM] features or the intersect command of your graphing utility to verify your answer.

a. Use a graphing utility (and the change-of-base property) to graph \(y=\log _{3} x\) b. Graph \(y=2+\log _{3} x, y=\log _{3}(x+2),\) and \(y=-\log _{3} x\) in the same viewing rectangle as \(y=\log _{3} x .\) Then describe the change or changes that need to be made to the graph of \(y=\log _{3} x\) to obtain each of these three graphs.
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