91Ó°ÊÓ

Logarithms are incredibly useful when solving equations, especially when they involve products, quotients, or powers. The laws of logarithms simplify expressions, making it easier to manipulate and solve equations. In the original exercise, the quotient rule is a key player. This rule states that

• \( \log_b \frac{M}{N} = \log_b M - \log_b N \)

When you have a subtraction of two logarithms with the same base, you can combine them into a single log using division. In our specific problem, we combined

• \( \log_5(2x+1) - \log_5(x-2) \)

into

• \( \log_5 \left(\frac{2x+1}{x-2}\right) \)

Understanding these laws not only makes problems easier to handle but also provides a deeper insight into the properties of logarithms. Familiarize yourself with the quotient, product, and power rules to tackle a range of logarithmic problems effectively.

Converting a logarithmic equation into exponential form is a technique that bridges the relationship between logs and exponents. This transformation is vital in simplifying and solving equations. To understand it, remember that if

• \( \log_b A = C \)

then

• \( b^C = A \)

This form allows us to eliminate the logarithm, making the equation easier to manage. In the exercise, we took the equation

• \( \log_5 \left( \frac{2x+1}{x-2} \right) = 1 \)

and rewrote it as

• \( 5^1 = \frac{2x+1}{x-2} \)

This conversion is crucial as it transforms a logarithmic relationship into an algebraic one, providing a straightforward path to finding the unknown variable. Mastering this concept will significantly enhance your problem-solving skills in mathematics.

Solving an equation involves isolating the variable. After converting a logarithmic equation to exponential form, the goal is to find the value of the unknown. In our exercise, we started with

• \( 5 = \frac{2x+1}{x-2} \)

First, eliminate the fraction by multiplying both sides by

• \((x-2) \)

This results in

• \( 5(x-2) = 2x + 1 \)

Distribute the 5 on the left side to get

• \( 5x - 10 = 2x + 1 \)

Next, bring like terms on one side by subtracting 2x from both sides:

• \( 3x - 10 = 1 \)

Add 10 to both sides:

• \( 3x = 11 \)

Finally, divide by 3:

• \( x = \frac{11}{3} \)

This process of isolating the variable step-by-step ensures you arrive at the correct solution. By practicing these procedures, you'll become proficient in solving a wide array of equations efficiently.