How do you find the required annual interest rate, to the nearest tenth of a percent, for $898 to grow to $1511 if interest is compounded monthly for 2 years?

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Answer 1 · 2015-10-17T00:09:16

Set up equation, then solve for i.

$898 {(1 + \frac{i}{12})}^{12 \cdot 2} = 1511$ $898(1+i/12)^(12*2)=1511$

${(1 + \frac{i}{12})}^{24} = \frac{1511}{898}$ $(1+i/12)^24=1511/898$

Now exponentiate each side of the equation by $\frac{1}{24}$ $1/24$

${[{(1 + \frac{i}{12})}^{24}]}^{\frac{1}{24}} = {(\frac{1511}{898})}^{\frac{1}{24}}$ $[(1+i/12)^24]^(1/24)=(1511/898)^(1/24)$

$(1 + \frac{i}{12}) = {(\frac{1511}{898})}^{\frac{1}{24}}$ $(1+i/12)=(1511/898)^(1/24)$

Now solve for i:

$i = [{(\frac{1511}{898})}^{\frac{1}{24}} - 1] \cdot 12 = 0.263 or 26.3 %$ $i=[(1511/898)^(1/24) -1]*12=0.263 or 26.3%$

hope that helped