APPLICATIONS OF EXPONENTIAL AND
LOGARITHMIC FUNCTIONS

APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS

(Interest Rate Word Problems)

1. To solve an exponential or logarithmic word problems, convert the narrative to an equation and solve the equation.

Problem 3: If you invested $1,000 in an account paying an annual percentage rate (quoted rate) compound daily (based on a bank year of 360 days) and you wanted to have $2,500 in your account at the end of your investment time, what interest rate would you need if the investment time were 1 year, 10 years, 20 years, 100 years?

Answer: 1 year = 91.75%, 10 years = 9.16%, 20 years = 4.58%, and 100 years 0.92%

Solution and Explanations:
Use the formula

displaymath195

where $2,500 is the balance at the end of a certain time period, $1,000 is the beginning investment, t is the number of years, and r is the annual percentage rate. The annual rate of r% is converted to a daily interest rate since the compounding is daily (360 times per year). Take the annual interest rate of 4% and divide by 360 to obtain the daily interest rate. The exponent is 360t because there are 360 compounding periods in every year. Therefore, 360t represents the number of compounding periods during t years.

To find the rate for 1 year:

Step 1: Substitute 1 for t in the equation

displaymath195

to derive r:

displaymath199

Step 2: Divide both sides of the above equation by $1,000:

displaymath201

Step 3: Take the natural logarithm of both sides of the above equation:

displaymath203

Step 4: Simplify the right side of the above equation using the third rule of logarithms:

displaymath205

Step 5: Divide both sides of the above equation by 360:

displaymath207

Step 6: Simplify the left side of the above equation:

displaymath209

Step 7: Convert the above equation to an exponential equation with base e and exponent 0.00254525203298:

displaymath211

Step 8: Simplify the left side of the above equation:

displaymath213

Step 9: Subtract 1 from both sides of the above equation:

displaymath215

Step 10: Multiply both sides of the above equation by 360:

displaymath217

This means the interest rate would have to be 91.75% (rounded) for the year.

Check:

displaymath219

This is a close enough check. Remember it will not check perfectly before we rounded the interest rate.

To find the rate for 10 year:

Step 1: Substitute 10 for t in the equation

displaymath195

to derive r:

displaymath223

Step 2: Divide both sides of the above equation by $1,000:

displaymath225

Step 3: Take the natural logarithm of both sides of the above equation:

displaymath227

Step 4: Simplify the right side of the above equation using the third rule of logarithms:

displaymath229

Step 5: Divide both sides of the above equation by 3600:

displaymath231

Step 6: Simplify the left side of the above equation:

displaymath233

Step 7: Convert the above equation to an exponential equation with base e and exponent 0.000254525203298:

displaymath235

Step 8: Simplify the left side of the above equation:

displaymath237

Step 9: Subtract 1 from both sides of the above equation:

displaymath239

Step 10: Multiply both sides of the above equation by 360:

displaymath241

This means the interest rate would have to be 9.16% (rounded) per year for 10 years.

Check:

displaymath243

This is a close enough check. Remember it will not check perfectly before we rounded the interest rate.

To find the rate for 20 year:

Step 1: Substitute 20 for t in the equation

displaymath195

to derive r:

displaymath247

Step 2: Divide both sides of the above equation by $1,000:

displaymath249

Step 3: Take the natural logarithm of both sides of the above equation:

displaymath251

Step 4: Simplify the right side of the above equation using the third rule of logarithms:

displaymath253

Step 5: Divide both sides of the above equation by 7200:

displaymath255

Step 6: Simplify the left side of the above equation:

displaymath257

Step 7: Convert the above equation to an exponential equation with base e and exponent 0.000127262601649:

displaymath259

Step 8: Simplify the left side of the above equation:

displaymath261

Step 9: Subtract 1 from both sides of the above equation:

displaymath263

Step 10: Multiply both sides of the above equation by 360:

displaymath265

This means the interest rate would have to be 4.58% (rounded) per year for 20 years.

Check:

displaymath267

This is a close enough check. Remember it will not check perfectly before we rounded the interest rate.

To find the rate for 100 year:

Step 1: Substitute 100 for t in the equation

displaymath195

to derive r:

displaymath271

Step 2: Divide both sides of the above equation by $1,000:

displaymath273

Step 3: Take the natural logarithm of both sides of the above equation:

displaymath275

Step 4: Simplify the right side of the above equation using the third rule of logarithms:

displaymath277

Step 5: Divide both sides of the above equation by 36000:

displaymath279

Step 6: Simplify the left side of the above equation:

displaymath281

Step 7: Convert the above equation to an exponential equation with base e and exponent 0.0000254525203298:

displaymath283

Step 8: Simplify the left side of the above equation:

displaymath285

Step 9: Subtract 1 from both sides of the above equation:

displaymath287

Step 10: Multiply both sides of the above equation by 360:

displaymath289

This means the interest rate would have to be 0.92% (rounded) per year for 100 years.

Check:

displaymath291

This is a close enough check. Remember it will not check perfectly before we rounded the interest rate.

If you would like to work another problem and check the answer and solution, click on Problem.

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Author: Nancy Marcus

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