College algebra problems on logarithmic and exponential functions are presented along with answers[cite: 1]. Each solution is hidden inside a collapsible dropdown so you can practice independently before reviewing the step-by-step calculations.
Let the logarithmic function \( f \) be defined by \( f(x) = 2\ln(2x - 1) \).
Let the exponential function \( h \) be defined by \( h(x) = 2 + e^x \).
The population of city A changes according to the exponential function \[ A(t) = 2.9 \times 2^{0.11t} \text{ (millions)} \] and the population of city B changes according to the exponential function \[ B(t) = 1.7 \times 2^{0.17t} \text{ (millions)} \] where \( t = 0 \) corresponds to 2009.
Find the inverse of the logarithmic function \( f \) defined by \( f(x) = 2 \log_5(2x - 8) + 3 \).
Solve the equation \( x = 2 \log_5(2y - 8) + 3 \) for \( y \) to obtain the inverse function:
\[ f^{-1}(x) = \dfrac{1}{2} 5^{\dfrac{x-3}{2}} + 4 \]Find the inverse of the exponential function \( h \) defined by \( h(x) = -2 \cdot 3^{-3x + 9} - 4 \).
Solve the equation \( x = -2 \cdot 3^{-3y + 9} - 4 \) for \( y \) to obtain the inverse function:
\[ h^{-1}(x) = \dfrac{-1}{3} \log_3 \left( \dfrac{x+4}{-2} \right) + 3 \]Solve the logarithmic equation defined by: \[ \ln(2x - 2) + \ln(4x - 3) = 2 \ln(2x) \]
Rewrite the given equation using logarithm properties:
\[ \ln\left[(2x - 2)(4x - 3)\right] = \ln(2x)^2 \]This gives the algebraic equation:
\[ (2x - 2)(4x - 3) = (2x)^2 \]Solve the quadratic equation for \( x \):
\[ x = 3 \quad \text{and} \quad x = \dfrac{1}{2} \]Checking the two values, only \( x = 3 \) is a valid solution to the given logarithmic equation (since \( x = \frac{1}{2} \) results in the logarithm of zero or a negative number).
\( A \), \( B \) and \( k \) in the exponential function \( f \) given by \[ f(x) = Ae^{kx} + B \] are constants. Find \( A \), \( B \) and \( k \) if \( f(0) = 1 \), \( f(1) = 2 \), and the graph of \( f \) has a horizontal asymptote \( y = -4 \).