SOLUTION: Thank you for your help! 2. $2,350 is invested in account paying 9% compounded semiannually. How much will the account be worth after 8 years? Show your work. 3. Cha

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Thank you for your help! 2. $2,350 is invested in account paying 9% compounded semiannually. How much will the account be worth after 8 years? Show your work. 3. Cha      Log On


   



Question 638021: Thank you for your help!
2. $2,350 is invested in account paying 9% compounded semiannually. How much will the account be worth after 8 years? Show your work.



3. Change each exponential form to an equivalent logarithmic form:
a. 32 = 25
b. = 25–3/2
c. 12 = 1441/2

4. Change each logarithmic form to an equivalent exponential form:
a. log 3 81 = 4
b. log 10 0.0001 = –4
c.

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
2. $2,350 is invested in account paying 9% compounded semiannually. How much will the account be worth after 8 years? Show your work
..
compound interest formula: A=P(1+i)^n, P=initial investment, i=interest rate/period, n=no. of periods, A=amt after n periods.
For given problem:
P=$2350,
i=.09/2=.045,
n=8*2=16
A=2350(1+.045)^16
=2350(1.045)^16
≈4753
after 8 years the account will be worth $4753
..
3. Change each exponential form to an equivalent logarithmic form:
These expressions are not in exponential form. See the definition of exponential forms in following problem 4.
a. 32 = 25
b. = 25–3/2
c. 12 = 1441/2
..
4. Change each logarithmic form to an equivalent exponential form:
a. log 3 81 = 4
Exponential form: 3^4=81 [base(3) raised to the log of the number(4)=number(81)]
..
b. log 10 0.0001 = –4
Exponential form: 10^-4=0.0001 [base(10) raised to the log of the number(-4)=number(0.0001)]