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# Solving for Time

Often we are interested in how long it will take to accumulate money or how long we’d need to extend a loan to bring payments down to a reasonable level. Note: This section assumes you’ve covered solving exponential equations using logarithms, either in prior classes or in the growth models chapter.

### Example 15

If you invest $2000 at 6% compounded monthly, how long will it take the account to double in value? This is a compound interest problem, since we are depositing money once and allowing it to grow. In this problem, P0 =$2000                  the initial deposit r = 0.06                       6% annual rate k = 12                          12 months in 1 year So our general equation is${{P}_{N}}=2000{{\left(1+\frac{0.06}{12}\right)}^{N\times12}}$. We also know that we want our ending amount to be double of $2000, which is$4000, so we’re looking for N so that PN = 4000. To solve this, we set our equation for PN equal to 4000. $4000=2000{{\left(1+\frac{0.06}{12}\right)}^{N\times12}}$              Divide both sides by 2000 $2={{\left(1.005\right)}^{12N}}$                                   To solve for the exponent, take the log of both sides $\log\left(2\right)=\log\left({{\left(1.005\right)}^{12N}}\right)$             Use the exponent property of logs on the right side $\log\left(2\right)=12N\log\left(1.005\right)$                     Now we can divide both sides by 12log(1.005) $\frac{\log\left(2\right)}{12\log\left(1.005\right)}=N$                              Approximating this to a decimal N = 11.581 It will take about 11.581 years for the account to double in value. Note that your answer may come out slightly differently if you had evaluated the logs to decimals and rounded during your calculations, but your answer should be close. For example if you rounded log(2) to 0.301 and log(1.005) to 0.00217, then your final answer would have been about 11.577 years.

### Try it Now Answers continued

4. d = unknown r = 0.16                       16% annual rate k = 12                         since we’re making monthly payments N = 2                           2 years to repay P0 = 3,000                   we’re starting with a $3,000 loan [latex-display]3,000=\frac{d\left(1-{{\left(1+\frac{0.16}{12}\right)}^{-2\times12}}\right)}{\frac{0.16}{12}}[/latex-display] Solving for d gives$146.89 as monthly payments. In total, she will pay $3,525.36 to the store, meaning she will pay$525.36 in interest over the two years. 5. 6. d = $30 The monthly payments r = 0.12 12% annual rate k = 12 since we’re making monthly payments P0 = 1,000 we’re starting with a$1,000 loan We are solving for N, the time to pay off the loan [latex-display]1,000=\frac{30\left(1-{{\left(1+\frac{0.12}{12}\right)}^{-N(12)}}\right)}{\frac{0.12}{12}}[/latex-display] Solving for N gives 3.396. It will take about 3.4 years to pay off the purchase.