Logarithmic and exponential functions — practice questions

First, we combine the logarithms on the left-hand side using the laws of logs.

1. Apply the power rule to the first term: $2\log_3 x = \log_3(x^2)$

The equation becomes: $\log_3(x^2) - \log_3(x-2) = 2$ 2. Apply the quotient rule: $\log_3\left(\frac{x^2}{x-2}\right) = 2$ 3. Convert the logarithmic equation to its equivalent exponential form. Remember, $\log_a b = c \iff a^c = b$. $\frac{x^2}{x-2} = 3^2$ $\frac{x^2}{x-2} = 9$ 4. Solve the resulting algebraic equation: $x^2 = 9(x-2)$ $x^2 = 9x - 18$ $x^2 - 9x + 18 = 0$ 5. Factorise the quadratic: $(x-3)(x-6) = 0$ This gives two potential solutions: $x=3$ or $x=6$. 6. Check for validity. The argument of a logarithm must be positive.

• For $x=3$: $\log_3(3)$ is valid. $\log_3(3-2) = \log_3(1)$ is valid.
• For $x=6$: $\log_3(6)$ is valid. $\log_3(6-2) = \log_3(4)$ is valid.

Both solutions are valid. Therefore, the solutions are $x=3$ and $x=6$.

(a) The initial temperature occurs at $t=0$. $T = 20 + 60e^{-0.05 \times 0} = 20 + 60e^0 = 20 + 60(1) = 80^{\circ}$C.

(b) The temperature of the surroundings is the value that $T$ approaches as time goes to infinity ($t \to \infty$). As $t \to \infty$, the term $e^{-0.05t}$ approaches 0. So, $T \to 20 + 60(0) = 20^{\circ}$C. This is the constant term in the equation.

(c) We need to find $t$ when $T=50$. $50 = 20 + 60e^{-0.05t}$

Isolate the exponential term: $30 = 60e^{-0.05t}$ $0.5 = e^{-0.05t}$

Take natural logarithms of both sides to solve for the exponent: $\ln(0.5) = \ln(e^{-0.05t})$ $\ln(0.5) = -0.05t$

Solve for $t$: $t = \frac{\ln(0.5)}{-0.05}$ $t \approx \frac{-0.693147}{-0.05} \approx 13.8629...$

To 3 significant figures, the time taken is 13.9 minutes.