7. A website decreases the price af some products by a certain percentage each day until someone buys the product. Winston is interested in buying a computer. He has R4 500 to spend, but the computer he wants is R7 200 . The price is to be decrease by \( 5 \% \) each day. How many days will it take before Winston can afford the computer?

Let the price of the computer on day \( n \) be given by \[ P(n) = 7200 \times 0.95^n. \] Winston can afford the computer when \[ P(n) \leq 4500. \] Substitute the expression for \( P(n) \): \[ 7200 \times 0.95^n \leq 4500. \] Divide both sides by \( 7200 \) (which is positive so the inequality sign stays the same): \[ 0.95^n \leq \frac{4500}{7200} = 0.625. \] Take the natural logarithm of both sides: \[ \ln(0.95^n) \leq \ln(0.625). \] By the logarithmic property, \( \ln(0.95^n) = n\ln(0.95) \), so we have: \[ n \ln(0.95) \leq \ln(0.625). \] Since \( \ln(0.95) \) is negative, dividing both sides by \( \ln(0.95) \) reverses the inequality: \[ n \geq \frac{\ln(0.625)}{\ln(0.95)}. \] Now, calculate the logarithms: \[ \ln(0.625) \approx -0.470003 \quad \text{and} \quad \ln(0.95) \approx -0.051293. \] Thus, \[ n \geq \frac{-0.470003}{-0.051293} \approx 9.165. \] Since \( n \) must be an integer and the price decreases at the end of each day, Winston must wait until the full day has passed. Therefore, it takes \( 10 \) days. The answer is: \[ \boxed{10} \]