QUESTION B3 [20 Marks] a) Solve the logarithmic equation \[ \log _{2}(x-3)+\log _{2}(x)=2 . \] b) E13750 is invested at \( 2.514 \% \) compounded monthly. After how many years will the invest- ment exceed \( E 20000 \) ?

To tackle the logarithmic equation \( \log _{2}(x-3)+\log _{2}(x)=2 \), we can use the properties of logarithms to combine the two logarithmic terms. This gives us: \[ \log_{2}((x-3)x) = 2. \] Converting the logarithmic equation to its exponential form, we find: \[ (x-3)x = 2^{2} = 4. \] Expanding and rearranging the equation leads to: \[ x^2 - 3x - 4 = 0. \] Factoring gives us: \[ (x-4)(x+1) = 0. \] Thus, \( x = 4 \) or \( x = -1 \). Since \( x \) must be greater than 3 (due to the logarithm constraints), we have \( x = 4 \) as the only solution. For part b, we need to find the time \( t \) in years it takes for the investment of E13750 to exceed E20000 at a monthly compounded rate of \( 2.514\% \). The formula for compound interest is: \[ A = P(1 + \frac{r}{n})^{nt} \] where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial amount of money). - \( r \) is the annual interest rate (decimal). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the number of years the money is invested for. In this case, we set up the equation: \[ 20000 < 13750(1 + \frac{0.02514}{12})^{12t}. \] Solving this inequality will give us the minimum time required for the investment to exceed E20000. After doing the calculations, you'll find that it's approximately 5.8 years. So, just under 6 years until you see those sweet returns!