1.5 Logarithms

Logarithms are the inverses of exponents. For \(a^x = b\), \(\log_a b = x\).

Laws of Logarithms:

• \(\log(ab) = \log a + \log b\) (Product Rule)
• \(\log\left(\frac{a}{b}\right) = \log a - \log b\) (Quotient Rule)
• \(\log(a^n) = n\log a\) (Power Rule)

Example 1: Simplify \(\log\left(\frac{28}{3}\right)\) given \(\log 2 = x\), \(\log 3 = y\), and \(\log 7 = z\).

Solution:

Step 1: Apply the logarithm property for division: \[ \log\left(\frac{28}{3}\right) = \log 28 - \log 3. \] Step 2: Break down 28 into factors: \[ \log 28 = \log(4 \cdot 7) = \log 4 + \log 7. \] Step 3: Substitute and simplify: \[ \log\left(\frac{28}{3}\right) = (\log 4 + \log 7) - \log 3. \] Step 4: Replace \(\log 4\) with \(2\log 2\): \[ \log\left(\frac{28}{3}\right) = (2\log 2 + \log 7) - \log 3. \] Therefore, \(\log\left(\frac{28}{3}\right) = 2x + z - y\).

Example 2: Evaluate \(x \log_2 8\) given \(\log_x 64 = 3\).

Solution:

Step 1: Express \(\log_x 64 = 3\) in exponential form: \[ x^3 = 64 \implies x = 4. \] Step 2: Evaluate \(\log_2 8\): \[ \log_2 8 = \log_2(2^3) = 3. \] Step 3: Multiply \(x\) by \(\log_2 8\): \[ x \log_2 8 = 4 \times 3 = 12. \] Therefore, \(x \log_2 8 = 12\).

Example 3: Solve \(\log_3 27 = 2x + 1\) to find the value of \(x\).

Solution:

Step 1: Simplify \(\log_3 27\): \[ \log_3 27 = \log_3(3^3) = 3. \] Step 2: Substitute into the equation: \[ 3 = 2x + 1. \] Step 3: Solve for \(x\): \[ 2x = 3 - 1 = 2. \] \[ x = \frac{2}{2} = 1. \] Therefore, \(x = 1\).