Problem 69 - Introductory and Intermediate Algebra for College Students 4th

Question

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. $$\log _{2}(x+2)-\log _{2}(x-5)=3$$

Step-by-Step Solution

Answer

The solution to the logarithmic equation \(\log _{2}(x+2)-\log _{2}(x-5)=3\) is \(x = 6\). This value is within the domain of the original expression.

Step 1: Use the properties of logarithms

The given equation is \( \log _{2}(x+2)-\log _{2}(x-5)=3 \). We can use the properties of logarithms to simplify the equation: \( \log _{2}(x+2) - \log _{2}(x-5) \) can be written as \( \log _{2}(\frac{x+2}{x-5}) \) by applying the quotient rule of logarithms which states that \( \log _{b}(M) - \log _{b}(N) = \log _{b}(\frac{M}{N}) \). Therefore, the equation becomes \( \log _{2}(\frac{x+2}{x-5})=3 \)

Step 2: Convert the logarithmic equation to an exponential equation

Rewrite the equation from logarithmic form to exponential form: from \( \log _{2}(\frac{x+2}{x-5})=3 \), we get \( 2^3 = \frac{x+2}{x-5} \) by using the basic definition of a logarithm which states that if \( \log _{b}(N)=p \), then \( b^p = N \)

Step 3: Solve the resulting equation

Now, solving the equation from Step 2, \( 2^3 = \frac{x+2}{x-5} \), we get \( 8=\frac{x+2}{x-5} \). After cross multiplying, the equation becomes \( 8x - 40 = x + 2 \). Rearranging the equation for \(x\), we obtain \( 8x - x = 40 + 2 \) thus \( 7x = 42 \). Solving for \(x\), we get \( x = \frac{42}{7} = 6 \)

Step 4: Check that the solution is within the domain

We must ensure that \(x=6\) satisfies the original logarithmic expressions. Thus the value of \(x\) must make \(x + 2 > 0\) and \(x - 5 > 0\). By substituting \(x = 6\) into these expressions, we find \(6 + 2 > 0\) and \(6 - 5 > 0\) which are both true, so \(x = 6\) is indeed within the domain of the original expressions and it is the solution.