Exponential function: For a > 0, the exponential function with base a is defined by. f(x) = a x

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1 MATH EXPONENTIAL FUNCTIONS KSU AND THEIR APPLICATIONS Defiitios: Expoetial fuctio: For a > 0, the expoetial fuctio with base a is defied by fx) = a x Horizotal asymptote: The lie y = c is a horizotal asymptote of the fuctio f if fx) c as x or x. Properties of the graph of fx) = a x, a > 0 Domai is all real umbers. Rage is 0, ). Always crosses through the poit 0, 1). y = 0 is a horizotal asymptote. If a > 1, the the fuctio is icreasig; if 0 < a < 1, the the fuctio is decreasig. Importat Formulas: Expoetial fuctios are used i a variety of importat formulas. Compoud Iterest: is calculated by the formula At) = P At) = amout after t years P = pricipal r = iterest rate 1 + r ) t = umber of times the iterest is compouded per year t = umber of years Compouded Cotiuously Iterest: is calculated by the formula At) = P e rt At) = amout after t years P = pricipal r = iterest rate t = umber of years

2 Expoetial fuctios ad their applicatios, page 2 Expoetial Growth: of a populatio icreases accordig to the formula P t) = P 0 e rt P t) = populatio after time t P 0 = iitial populatio r = growth rate t = time Importat Properties: Every expoetial fuctio is a oe-to-oe fuctio ad hece has a iverse. Commo Mistakes to Avoid: Do NOT use the compouded cotiuously formula uless it says compouded cotiuously i the problem. I the expoetial growth ad compouded cotiuously formulas the rt is the expoet o e. Do NOT multiply e by rt. Remember to covert all iterest or growth rates to a decimal before substitutig ito a formula.

3 Expoetial fuctios ad their applicatios, page 3 PROBLEMS 1. If \$5, 000 is ivested at a rate of 8%, compouded weekly, fid the value of the ivestmet after 7 years. Here P = 5000, t = 7, r =.08 ad = 52 sice there are 52 weeks i a year. Substitutig these values ito our compoud iterest At) = P 1 + r ) t A7) = ) = ) 364 To fid the amout due at the ed of 8 years, we chage t = 8. At) = P 1 + r ) t A8) = ) = ) 32 = \$ due at the ed of 8 years = \$ If \$4000 is borrowed at a rate of 16% iterest per year, compouded quarterly, fid the amout due at the ed of 4 years? 8 years? Here P = 4000, r =.16 ad = 4 sice there are 4 quarters i a year. To fid the amout due at the ed of 4 years we let t = 4. Substitutig ito the compoud iterest At) = P 1 + r ) t A4) = ) = ) 16 = If \$3000 is borrowed at a rate of 12% iterest per year, fid the amout due at the ed of 5 years if the iterest is compouded aually? mothly? daily? For this problem, we have P = 3000, r =.12 ad t = 5. If the moey is compouded aually, the = 1. Substitutig ito the compoud iterest At) = P 1 + r ) t A5) = ) = ) 5 = \$ due if compouded aually \$ due at the ed of 4 years

4 Expoetial fuctios ad their applicatios, page 4 Whe the moey is compouded mothly, = 12 sice there are 12 moths i oe year. Therefore, At) = P 1 + r ) t A5) = ) = ) 60 = If \$3000 is borrowed at a rate of 12% iterest per year, fid the amout due at the ed of 5 years if the iterest is compouded cotiuously. For this problem, we use the compouded cotiuously formula with P = 3000, r =.12, ad t = 5. Substitutig everythig i, we get At) = P e rt A5) = 3000e.12 5 = ) \$ due if compouded mothly Fially, if the moey is compouded daily, the = 365 sice there are 365 days i oe year. Hece, = \$ due if compouded cotiuously At) = P 1 + r ) t A5) = ) = ) 1825 = \$ due if compouded daily 5. Fid the amout that must be ivested at % today i order to have \$100, 000 i 20 years if the ivestmet is compouded cotiuously. Here, we have A20) = 100, 000, r =.055, ad t = 20. Substitutig ito the compouded cotiuously iterest At) = P e rt 100, 000 = P e , 000 = P e , 000 e 1.1 = P = P \$33, must be ivested ow

5 Expoetial fuctios ad their applicatios, page 5 6. The populatio of a certai city has a relative growth rate of 9% per year. The populatio i 1978 was 24,000. Fid the projected populatio of the city for the year Sice 1978 is our startig date, 2010 refers to t = 22. Also, we kow that P 0 = 24, 000 ad r =.09. Substitutig ito our expoetial growth P t) = P 0 e rt P 22) = 24, 000e = 24, 000e 1.98 = , 825 people i The relative growth rate for a certai bacteria populatio is 75% per hour. A small culture is formed ad 4 hours later a cout shows approximately 32,500 bacteria i a culture. Fid the iitial umber of bacteria i the culture ad estimate the umber of bacteria 6 hours from the time the culture was started. To solve for the iitial umber of bacteria i the culture, we will use the expoetial growth formula with r =.75, t = 4, ad P 4) = 32, 500. Therefore, P t) = P 0 e rt 32, 500 = P 0 e , 500 = P 0 e 3 32, 500 e 3 = P = P bacteria iitially Sice we have the iitial populatio of 1618, substitutig this ito our expoetial growth formula with t = 6, we get P t) = P 0 e rt P 6) = 1618e.75 6 = 1618e 4.5 = , 647 bacteria after 6 hours

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