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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. $$\log (x+3)+\log (x-2)=\log 14$$

Short Answer

Expert verified
The solution to the given equation is \(x = 4.00\).

Step by step solution

01

Combine the Logarithmic Expressions

Based on the properties of logarithms, we know that the logarithm of a product is the sum of the logarithms. Therefore, the equation can be rewritten as: \(\log [(x+3)(x-2)] = \log 14\).
02

Remove the Logarithms

The equation has now taken the form \(\log_b m = \log_b n\), where \(b\), \(m\), and \(n\) are positive numbers and \(b \neq 1\). According to the equality property of logarithms, if two logarithms with the same base are equal, then their arguments are also equal. Therefore, we can remove the logarithms from our equation to get \((x+3)(x-2) = 14\).
03

Solve for \(x\)

We now have a quadratic equation. First, we expand it: \(x^2 + x - 6 = 14\). This simplifies to \(x^2 + x - 20 = 0\). By factoring, we get \((x-4)(x+5) = 0\). Setting each factor equal to zero gives us the potential solutions \(x = 4\) and \(x = -5\).
04

Check the Domain

We need to check if our solutions don't make the original logarithm expressions undefined. Testing \(x = 4\) in the original equation, we have \(\log (4+3) + \log (4-2)\) which gives \(\log 7 + \log 2 = \log 14\), as required. The value \(x = -5\) makes the term '\(x-2\)' negative, which is undefined for real number logarithms. Therefore, it must be rejected.
05

Approximate the Solution

The only solution which is in the domain of the original logarithmic expressions is \(x = 4\). In decimal form, this is already correct to two decimal places.

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Most popular questions from this chapter

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