91Ó°ÊÓ

The quotient rule of logarithms is a useful property that allows us to simplify logarithmic expressions involving division. The rule states: for any positive real numbers a and b, and any positive base b (where b≠1), the following holds: \[ \log_b\left(\frac{a}{c}\right) = \log_b(a) - \log_b(c)\] This means when you subtract logarithms with the same base, you can combine them into a single logarithm of the quotient of their arguments.
For example, if you have \( \log_4(x+3) - \log_4(x-5) = 2 \), you can use the quotient rule to rewrite it as \( \log_4\left(\frac{x+3}{x-5}\right) = 2 \). This step is crucial for solving logarithmic equations, as it condenses multiple logarithms into one, making the equation easier to handle and solve.
Examine the problem for understanding: \( \log_4(x+3) = 2 + \log_4(x-5) \). Begin by isolating the logarithmic terms on one side using subtraction: \( \log_4(x+3) - \log_4(x-5) = 2 \). Now, apply the quotient rule to get: \( \log_4\left(\frac{x+3}{x-5}\right) = 2 \).

An exponential equation is an equation in which a variable appears in the exponent. Solving it usually involves logarithms because they are the inverse of exponentials.
In the exercise, after applying the quotient rule, you get: \( \log_4\left(\frac{x+3}{x-5}\right) = 2 \). To solve for \( x \), convert this logarithmic equation to its exponential form: \( b^y = x \). Here: \( b = 4 \), \( y = 2 \), and \( x = \frac{x+3}{x-5} \). This gives: \( \frac{x+3}{x-5} = 4^2 = 16 \).
By rewriting the logarithmic equation in its exponential form, you are essentially undoing the logarithm to make the equation simpler and more solvable. This step is often critical in transitioning from a logarithmic context to an algebraic one.

Solving logarithmic equations involves several steps: simplifying the logarithmic terms, converting to exponential form, and solving the resulting algebraic equation. Let's go through these steps using the example provided.
Once we have \( \frac{x+3}{x-5} = 16 \), we need to get rid of the fraction to solve for \( x \). Multiply both sides by \( x-5 \): \( x + 3 = 16(x - 5) \).
Simplify and solve for \( x \):

• Expand: \( x + 3 = 16x - 80 \)
• Combine like terms: \( 3 + 80 =16x - x \)
• Simplify: \( 83 = 15x \)
• Divide by 15: \( x = \frac{83}{15} \)

Thus, \( x \approx 5.533 \).
Verify the solution by checking if the arguments in the original logarithms are valid. Substitute \( x \approx 5.533 \) back. The arguments are valid if they are greater than zero:

• \( x + 3 = 5.533 + 3 = 8.533 \)
• \( x - 5 = 5.533 - 5 = 0.533 \)

Both values fall within the logarithms' domain, confirming the solution is correct.