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Chapter 12: Problem 57

Solve. Where appropriate, include approximations to three decimal places. $$ \log _{7}(x+1)+\log _{7}(x+2)=\log _{7} 6 $$

Short Answer

Expert verified
The solution is x = 1.

Step by step solution

01

- Use Logarithm Properties

Recall the property of logarithms that states: \(\text{If } \log_{a}b + \log_{a}c = \log_{a}(bc)\). Apply this property to the given equation:\(\log_{7}(x+1) + \log_{7}(x+2) = \log_{7}(6)\)This simplifies to:\(\log_{7}((x + 1)(x + 2)) = \log_{7}(6)\)
02

- Equate the Arguments

Since the logs are equal, their arguments must be equal as well:\((x + 1)(x + 2) = 6\)
03

- Expand and Simplify

Expand the left-hand side of the equation:\(x^2 + 3x + 2 = 6\)Subtract 6 from both sides to set the equation to zero:\(x^2 + 3x + 2 - 6 = 0\)This simplifies to:\(x^2 + 3x - 4 = 0\)
04

- Factor the Quadratic Equation

Factor the quadratic equation:\(x^2 + 3x - 4 = (x + 4)(x - 1)\)
05

- Solve for x

Set each factor to zero and solve for x:\(x + 4 = 0\) or \(x - 1 = 0\)Thus, \(x = -4\) or \(x = 1\)
06

- Check Solutions

Check each solution in the context of the original logarithmic equation. Only positive arguments in the logarithms are valid:- If \(x = -4\), then \(\log_{7}(-4+1)\) and \(\log_{7}(-4+2)\) are undefined.- If \(x = 1\), then \(\log_{7}(1+1)\) and \(\log_{7}(1+2)\) are valid.Thus, only \(x = 1\) is valid.

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Key Concepts

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

logarithm properties
To solve logarithmic equations, it is crucial to understand the properties of logarithms. One essential property is:
If \(\log_{a}b + \log_{a}c = \log_{a}(bc)\)\.
This property allows you to combine two logarithms with the same base into a single logarithm.
For example, in the equation \( \log_{7}(x+1) + \log_{7}(x+2) = \log_{7}(6) \), you can combine the left-hand side to get:
\(\log_{7}((x + 1)(x + 2)) = \log_{7}(6)\).
This makes it easier to solve because now you have a single logarithm on each side.
quadratic equations
After using the properties of logarithms, we often need to solve a quadratic equation. In our example, combining the logarithms led to:
\( (x + 1)(x + 2) = 6 \).
Expanding and simplifying this gives us: \(\ x^2 + 3x - 4 = 0\).
This is a standard quadratic equation of the form \(ax^2 + bx + c = 0 \).
To solve it, you can factorize the equation as:
\(\ x^2 + 3x - 4 = (x + 4)(x - 1)\).
Setting each factor to zero yields two potential solutions: \( x + 4 = 0 \) or \( x - 1 = 0 \),
which means \ x = -4 \ or \ x = 1 \.
valid solutions in logarithmic equations
Not all mathematical solutions are valid in the context of logarithmic equations. For a solution to be valid, the argument of the logarithm must be positive.
In our example, after solving the quadratic equation, we get two potential solutions: \ x = -4 \ and \ x = 1 \.
  • If \ x = -4 \, then the arguments become \( \log_{7}(-4+1) \) and \ \log_{7}(-4+2) \, which are undefined because logarithms of negative numbers do not exist.
  • If \ x = 1 \, the arguments are \( \log_{7}(2) \) and \ \log_{7}(3) \, which are both valid because they are positive numbers.

Therefore, the only valid solution in the given context is \( x = 1 \).
Always remember to check the constraints of the logarithmic functions to discard any invalid solutions.

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Most popular questions from this chapter

As of February \(2010,\) Alex Rodriguez of the New York Yankees had the largest contract in sports history. As part of the 10 -year \(\$ 275\) -million deal, he will receive \(\$ 20\) million in \(2016 .\) How much money would need to be invested in 2008 at \(4 \%\) interest, compounded continuously, in order to have \(\$ 20\) million for Rodriguez in \(2016 ?\) (This is much like determining what \(\$ 20\) million in 2016 is worth in 2008 dollars.

Solve. Value of a Sports Card. Legend has it that because he objected to teenagers smoking, and because his first baseball card was issued in cigarette packs, the great shortstop Honus Wagner halted production of his card before many were produced. One of these cards was purchased in 1991 by hockey great Wayne Gretzky (and a partner) for \(\$ 451,000 .\) The same card was sold in 2007 for \(\$ 2.8\) million. For the following questions, assume that the card's value increases exponentially, as it has for many years. a) Find the exponential growth rate \(k,\) and determine an exponential function that can be used to estimate the dollar value, \(V(t),\) of the card \(t\) years after 1991 b) Predict the value of the card in 2012 c) What is the doubling time for the value of the card? d) In what year will the value of the card first exceed \(\$ 4,000,000 ?\)

Use the properties of logarithms to find each of the following. $$\log _{5}(125 \cdot 625)$$

$$\text { If } \log _{a} x=2, \text { what is } \log _{1 / a} x ?$$

Solve for \(x\) $$ \frac{3.01}{\ln x}=\frac{28}{4.31} $$
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