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Exercise 5.4 · Q17

Q.A stock begins to pay dividends with the first dividend, one year from now, expected to be ₹100. Each year the dividend is 10% larger than the previous year's dividend. In what year is the dividend paid larger than ₹1000? (Use the concept of logarithm to solve the question)

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Model the dividend as a GP Dn=100(1.10)n−1D_n=100(1.10)^{n-1} and solve Dn>1000D_n>1000 for the smallest integer year nn using logarithms.

nn-th term of a GP: Dn=D1⋅r n−1D_n=D_1\cdot r^{\,n-1}, where D1=D_1= first dividend, r=1+r=1+growth rate. To solve rk>Nr^{k}>N, take logs: k>log⁡rN=log⁡Nlog⁡rk>\log_r N=\dfrac{\log N}{\log r}.

  1. First dividend (Year 1, one year from now): D1=₹100D_1=\text{₹}100.
  2. Each year's dividend is 10%10\% larger than the previous, so common ratio r=1.10r=1.10.
  3. Dividend in Year nn: Dn=100(1.10)n−1D_n=100(1.10)^{n-1}.
  4. Require Dn>1000D_n>1000:

100(1.10)n−1>1000 ⟹ (1.10)n−1>10100(1.10)^{n-1}>1000\ \Longrightarrow\ (1.10)^{n-1}>10

  1. Take log⁡10\log_{10} of both sides:

(n−1)log⁡(1.10)>log⁡(10)(n-1)\log(1.10)>\log(10)

  1. Solve for n−1n-1: n−1>log⁡10log⁡1.10=10.041393≈24.16n-1>\dfrac{\log 10}{\log 1.10}=\dfrac{1}{0.041393}\approx24.16 …

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