AN ALTERNATIVE TO THE LOAN LENGTH FORMULA

equation.

y = ex

To find the value of x when given a particular y, use the following algebraic transformation.

x = ln y

AN

?ALTERNATIVE

?

TO

?THE

?

LOAN

?logarithm

LENGTH

?ofFy¡±

ORMULA

?

This is read as ¡°x equals

the natural

or ¡°when (e p.189)

?

is

raised

to

the

exponent

x,

the

resulting

value

is

y.¡±

?

Before the use of calculators, people used a logarithm table to deterEXAMPLE

mine the2exponent values. Now, graphing calculators have a natural

logarithm

key

For$25,000

example,

following

equation.

Claude

wants

to[LN].

borrow

toconsider

purchasethe

a car.

After looking

at his

?

monthly130

budget,

= ex he realizes that all he can afford to pay per month is

$300. The bank is offering a 5.9% loan. What would the length of his

To solve

x, that

is, to

nd the

which you need to raise

loan

need for

to be

so that

hefican

stayexponent

within histobudget?

e to get 130, you need to use the [LN] key and then enter 130 into the

SOLUTION

To solve

problem, it is necessary

to perform some

calculator. The

resultthis

is approximately

4.9.

algebraic manipulations on the monthly loan payment formula.

The development of the loan length formula is beyond the scope of

To

find

the length

of the loan

giventhe

theuse

amount

the monthly

pay-in

this

course.

That formula

requires

of theofnatural

logarithm

ment,

you

need

to

solve

for

the

exponent

t.

To

solve

for

an

exponent,

order to solve for the exponent t.

you need to understand the concept of a natural logarithm. In

Loan

Formula

Lesson

3-6, Length

you learned

about the constant e. Examine the following

where M = monthly payment

equation. M

M

r

ln x__ ? ln __ ? ___

p = principal

p

12

y ______________________

=e p

r = interest rate

t=

r

___

To find the value

when

given a particular

use theoffollowing

alge12lnof1x+

t = y,

number

years

braic transformation. 12

( ) ( (

(

)

))

) ( (

(

))

x = ln y

300

300

0.059

ln _______ ? ln _______ ? ______

Substitute p = 25,000,

25,000

25,000

This

read and

as ¡°x

equals

the natural

logarithm of y¡± or ¡°when e is12

M =is300,

r=

0.059.

t = __________________________________

raised to the exponent x, the resulting value is y.¡±

0.059

12ln 1 + ______

12

Before the use of calculators, people used a logarithm table to determine

the exponent

values.

Now, graphing

Calculate

to the nearest

hundredth

of a year.calculators

t ¡Ö have

8.96 a natural

logarithm key [LN]. For example, consider the following equation.

Claude would need to take out a loan for about 9 years.

130 = ex

(

)

solve forYOUR

x, that UNDERSTANDING

is, to find the exponent to which you need to raise

¡öToCHECK

e In

to Example

get 130, you

need

to usewould

the [LN]

key and in

then

130 payment

into the

2, what

impact

an increase

theenter

monthly

calculator.

The

result

is

approximately

4.9.

of $50 have on the length of the loan?

The development of the loan length formula is beyond the scope of

this course. That formula requires the use of the natural logarithm in

order to solve for the exponent t.

is a lesson on how the natural log can be used in this

particular case. The resulting formula is complex and

again requires the proper

placement of parentheses.

CHECK YOUR

UNDERSTANDING

Answer The length of the

loan would drop to about

EXAMPLE

7.35 years if2the monthly

?

payment

is increased

by

Since

the variable

repre$50. time in the monthly

senting

payment

an

Studentsformula

need toisrecalcuexponent,

solving

for that

late the time

formula

from

exponent

the as

Examplerequires

2 using $350

use

logarithms.

SomeAsk

theofmonthly

payment.

students

already

have

studentsmay

what

they expect

been

introduced

to

logs

in a

will happen to the length

previous

math

course

while

of the loan before they actufor

others,

may be their

ally

do thethis

calculations.

first introduction to the

topic. This is not meant to

be a comprehensive lesson

on the use of logs. Rather, it

is a lesson on how the natural log can be used in this

particular case. The resulting formula is complex and

again requires the proper

placement of parentheses.

CHECK YOUR

UNDERSTANDING

Answer The length of the

loan would drop to about

7.35 years if the monthly

payment is increased by

$50.

Students need to recalculate the time formula from

Example 2 using $350 as

the monthly payment. Ask

students what they expect

will happen to the length

of

? the loan before they actually do the calculations.

?

Graphing

?

Calculator

?(or

?Graphing

?Software)

?

Loan Length Formula

4-3

Loan Calculations and Regression

?

where M = monthly payment

M

M

r

?

ln __ ? ln __ ? ___

p = principal

p

p

( ) ( (

))

12

t = ______________________

r

12ln 1 + ___

12

(

)

r = interest rate

t = number of years

!"#$%&'!&()'!&*+%,-,+$./01122234(+'5+6"

Substitute p = 25,000,

M = 300, and r = 0.059.

?

?

t=

189

(

) ( (

))

300

300

0.059

ln _______ ? ln _______ ? ______

25,000

25,000

12

__________________________________

?

?

?

?

?

?

?

?

?

Calculate to the nearest hundredth of a year.

(

0.059

12ln 1 + ______

12

)

?

t ¡Ö 8.96

Claude would need to take out a loan for about 9 years.

¡ö CHECK YOUR UNDERSTANDING

In Example 2, what impact would an increase in the monthly payment

of $50 have on the length of the loan?

'"7'87

?

SPREADSHEET

?

?

?

?

?

?

?

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