Math Test 3/6 Review Questions With Complete Solutions
Solve the exponential growth/decay problem:
A savings account balance is compounded
annually. If the interest rate is 2% per
year and the current balance is $1,754.00,
what will the balance be 9 years from
now?
1754×1.02^9=2,096.19
y=b^0
1
y=b^1
b
HA:
y=0
b>1
growth
0<b<1:
decay
f(x)=a.b^x
a: starting amount
b=growth or decay factor
b=(1+%)+growth or decay
, 1: initial value
percent divided by 100 before added
Y=Pe^rt, continuously compounded
p=start
e=2nd divide in calc.
r=%
t=time
If b^p=n then
logb(N)=P where b>0, b isn't 1, and n>0
Logb(N)=P
b-base
N-what it is equal to
P-exponent
b^P=N is
exponential form
logb(N)=P is
log form
When there's no number under log,
the common log base 10 is assumed
When In,
natural log e is assumed
Plug in answers to check for
extraneous solutions
Solve the exponential growth/decay problem:
A savings account balance is compounded
annually. If the interest rate is 2% per
year and the current balance is $1,754.00,
what will the balance be 9 years from
now?
1754×1.02^9=2,096.19
y=b^0
1
y=b^1
b
HA:
y=0
b>1
growth
0<b<1:
decay
f(x)=a.b^x
a: starting amount
b=growth or decay factor
b=(1+%)+growth or decay
, 1: initial value
percent divided by 100 before added
Y=Pe^rt, continuously compounded
p=start
e=2nd divide in calc.
r=%
t=time
If b^p=n then
logb(N)=P where b>0, b isn't 1, and n>0
Logb(N)=P
b-base
N-what it is equal to
P-exponent
b^P=N is
exponential form
logb(N)=P is
log form
When there's no number under log,
the common log base 10 is assumed
When In,
natural log e is assumed
Plug in answers to check for
extraneous solutions
