Logarithmic and exponential functions are fundamental to algebra. Because they are inverse functions, we can solve complex equations by converting between their forms. This guide provides comprehensive review rules and solved practice problems for Grade 11 math students.
Logarithmic and exponential expressions are equivalent:
\[ \log_b(x) = y \iff x = b^y \]Where \( b > 0 \), \( b \neq 1 \), and \( x > 0 \).
a) \( \log_x(a) = c \)
b) \( \log_b(2x + 1) = 3 \)
Using \( \log_b(x) = y \iff x = b^y \):
a) \( \mathbf{a = x^c} \)
b) \( \mathbf{2x + 1 = b^3} \)
a) \( 3^x = m \)
b) \( x^2 = a \)
Using \( x = b^y \iff y = \log_b(x) \):
a) \( \mathbf{x = \log_3(m)} \)
b) \( \mathbf{2 = \log_x(a)} \)
Find the value of: a) \( \log_2(16) \) | b) \( \log_3(27) \) | c) \( \log_{25}(5) \)
a) Let \( y = \log_2(16) \implies 2^y = 16 \). Since \( 16 = 2^4 \), \( y = \mathbf{4} \).
b) Let \( y = \log_3(27) \implies 3^y = 27 \). Since \( 27 = 3^3 \), \( y = \mathbf{3} \).
c) Let \( y = \log_{25}(5) \implies 25^y = 5 \). Since \( 5 = \sqrt{25} = 25^{1/2} \), \( y = \mathbf{1/2} \).
Solve for \( x \): \( \log_2(3x + 1) = 4 \)
Convert to exponential form:
\[ 3x + 1 = 2^4 \] \[ 3x + 1 = 16 \] \[ 3x = 15 \implies \mathbf{x = 5} \]Solve for \( x \): \( \log_4 \frac{x+1}{2x-1} = 0 \)
Convert to exponential form:
\[ \frac{x+1}{2x-1} = 4^0 = 1 \] \[ x + 1 = 2x - 1 \] \[ \mathbf{x = 2} \]Solve for \( x \): \( 4^{2x+1} = 16 \)
Write both sides with base 4:
\[ 4^{2x+1} = 4^2 \]Equate the exponents:
\[ 2x + 1 = 2 \implies 2x = 1 \implies \mathbf{x = 0.5} \]Solve: \( 2^{2x} - 6(2^x) = -8 \)
Note that \( 2^{2x} = (2^x)^2 \). Let \( u = 2^x \):
\[ u^2 - 6u + 8 = 0 \]Factor the quadratic:
\[ (u - 2)(u - 4) = 0 \implies u = 2, u = 4 \]Substitute back for \( x \):
Explore more Grade 12 Logarithm Questions or visit our High School Math Hub.