Question

If \(\log 3=P\) and \(\log 5=Q,\) write an algebraic expression in terms of \(P\) and \(Q\) for each of the following. a) \(\log \frac{3}{5}\) b) \(\log 15\) c) log \(3 \sqrt{5}\) d) \(\log \frac{25}{9}\)

Short answer

Apply the properties of logarithms. For each part: \ log\(\frac{3}{5}\) = P - Q\ \(\; \ \log 15\ \text{ is } \ \log(3*5)\ = P + Q\ \(\ \log 3\sqrt{5}= P + \frac{1}{2}Q\ \(\ \log\ \frac{25}{9}\ =2Q-2P.

Step-by-step solution

Understanding the problem

Given \(\log 3 = P\)\ and \(\log 5 = Q\), expressions involving logarithms need to be rewritten using the properties of logarithms in terms of \ P \ and \ Q \.

Key concepts

Logarithm Properties

Logarithms have several properties that make them useful for solving various mathematical problems. The key properties include:

• Product Property: \(\text{log}_{b}(xy) = \text{log}_{b}(x) + \text{log}_{b}(y)\) - This property tells us that the logarithm of a product is the sum of the logarithms of the factors.
• Quotient Property: \(\text{log}_{b}(x/y) = \text{log}_{b}(x) - \text{log}_{b}(y)\) - This states that the logarithm of a quotient is the difference of the logarithms.
• Power Property: \(\text{log}_{b}(x^k) = k \text{log}_{b}(x)\) - This tells us that the logarithm of a number raised to an exponent is the exponent times the logarithm of the base number.

Understanding these properties helps simplify and manipulate logarithmic expressions.

Algebraic Manipulation

Let's use algebraic manipulation to solve each part of the given exercise by applying the logarithm properties.

• Part (a): \(\text{log} \frac{3}{5}\) - We use the Quotient Property here.
  \[ \text{log} \frac{3}{5} = \text{log} 3 - \text{log} 5 \]
  Given \(\text{log} 3 = P\) and \(\text{log} 5 = Q\), it simplifies to:
  \[ \text{log} \frac{3}{5} = P - Q \]
• Part (b): \(\text{log} 15\) - We use the Product Property.
  \[ \text{log} 15 = \text{log} (3 \times 5) = \text{log} 3 + \text{log} 5 \]
  Simplifies to:
  \[ \text{log} 15 = P + Q \]
• Part (c): \(\text{log} 3 \sqrt{5}\) - This involves both the Product and Power Properties.
  \[ \text{log} 3 \sqrt{5} = \text{log} 3 + \text{log} \sqrt{5} = \text{log} 3 + \frac{1}{2} \text{log} 5 \]
  Simplifies to:
  \[ \text{log} 3 \sqrt{5} = P + \frac{1}{2} Q \]
• Part (d): \(\text{log} \frac{25}{9}\) - Using both the Quotient and Power Properties.
  \[ \text{log} \frac{25}{9} = \text{log} 25 - \text{log} 9 \]
  Further simplifying using the Power Property:
  \[ \text{log} 25 = 2 \text{log} 5 = 2Q \quad \text{and} \quad \text{log} 9 = 2 \text{log} 3 = 2P \]
  Simplifies to:
  \[ \text{log} \frac{25}{9} = 2Q - 2P \]

These manipulations help represent expressions involving logarithms in terms of given variables P and Q.

Precalculus Problems

Precalculus often involves using logarithms to solve various types of problems.
Understanding and applying logarithm properties is essential for simplifying complex expressions.
Precalculus students frequently encounter exercises like the given one, where they translate expressions into simpler forms using given information.
Students must become comfortable with:

• Identifying which logarithm property to use
• Substituting given values to simplify expressions
• Combining multiple properties to tackle more complex problems

This solidifies their foundation for higher-level math where logarithms are used more extensively.

Logarithmic Identities

Logarithmic identities are crucial tools in algebra and calculus.
They help in simplifying and solving equations involving logarithms.
The identities most commonly used include:

• Change of Base Formula: \(\text{log}_{b}(x) = \frac{\text{log}_{a}(x)}{\text{log}_{a}(b)}\)
• Inverse Relationship: \(\text{log}_{b}(b^{x}) = x\) and \ (b^{\text{log}_{b}(x)} = x) \

By understanding and mastering these identities, students can approach logarithmic problems with confidence.
These identities also play a crucial role in the transition from algebraic manipulation to calculus.