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The number of years n required for an investment at interest rate r to double in value must satisfy (1 + r)n = 2. Using ln 2 = .69 and the approximation ln(1 + r) ≈ r valid for small r, show that n ≈ 69/i, where i is the interest rate percentage (that is, i = 100r). Using the better approximation ln(1+r) ≈ r − 1 2 r2, show that for r ≈ .08 there holds n ≈ 72/i.

Respuesta :

zainsubhani

Answer:

Showing n=69/i :

n*r=0.69

where r=i/100

n*(i/100)=0.69

Solving the above Equation:

n=69/i  (Proved)

Showing n=72/i :

[tex]n*(0.08-\frac{1}{2}(0.08)^2)=0.69\\n* 0.0768=0.69\\n=8.98[/tex]

Above we calculated n=8.98 ≈ 9 (Proved n=72/i)

Explanation:

Given:

[tex](1+r)^n=2[/tex]

ln(2)=0.69

i=100*r means r=i/100

Solution:

Showing n=69/i :

[tex](1+r)^n=2[/tex]

Taking ln on both sides:

[tex]ln(1+r)^n=ln (2)\\n*ln(1+r)=0.69[/tex]

From given data ln(1+r) ≈ r

Above Equation will become:

n*r=0.69

where r=i/100

n*(i/100)=0.69

Solving the above Equation:

n=69/i  (Proved)

Showing n=72/i :

As we know i=100*r

when r=0.08,

i=100*0.08=8

[tex]n=72/i =72/8 =9[/tex]

Now:

[tex](1+r)^n=2[/tex]

Taking ln on both sides:

[tex]ln(1+r)^n=ln (2)\\n*ln(1+r)=0.69[/tex]

From given data ln(1+r)≈[tex]r-\frac{1}{2} r^2[/tex]

Above Equation will become:

[tex]n*r-\frac{1}{2} r^2=0.69[/tex]

where r=0.08, Now:

[tex]n*(0.08-\frac{1}{2}(0.08)^2)=0.69\\n* 0.0768=0.69\\n=8.98[/tex]

Above we calculated n=8.98 ≈ 9 (Proved n=72/i)

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