SOLUTION: An investment of $25,400 is placed into an account that earns 6.5% interest compounded quarterly. In how many years will the investment be worth twice the original amount? 2)

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: An investment of $25,400 is placed into an account that earns 6.5% interest compounded quarterly. In how many years will the investment be worth twice the original amount? 2)       Log On


   

Question 633098: An investment of $25,400 is placed into an account that earns 6.5% interest compounded quarterly. In how many years will the investment be worth twice the original amount?
2) Solve: log3(x+8) + log3x=2 The 3 are are at the bottom right corner.
3) log5x-log5(5-2)=log5(4) The 5 is in the bottom right corner also
4) 2^3x+1=13.4
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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An investment of $25,400 is placed into an account that earns 6.5% interest
compounded quarterly.
In how many years will the investment be worth twice the original amount?
:
How long to double, the amt doesn't matter
let t = no. of yrs to double
(1+.065%2F4)^(4t) = 2
%281.01625%29%5E%284t%29 = 2
4t = log%282%29%2Flog%281.01625%29
4t = 43
t = 43/4
t = 10.75 yrs to double in value
:
:
2) Solve: log3(x+8) + log3(x) = 2
addition of logs is multiply
log3(x(x+8)) = 2
exponent of equiv of logs
x(x+8) = 3^2
x^2 + 8x = 9
a quadratic equation
x^2 + 8x - 9 = 0
Factors to
(x+9)(x-1) = 0
only the positive solution can be used here
x = 1
:
:
3) log5(x) - log5(5-2)= log5(4)
log5(x) - log5(3)= log5(4)
subtraction of log means divide
log5%28x%2F3%29 = log5(4)
eliminate log5 from both sides
x%2F3 = 4
x = 3(4)
x = 12
:
:
4) 2^3x+1=13.4
Assume this is
2%5E%28%283x%2B1%29%29 = 13.4
3x + 1 = log%2813.4%29%2Flog%282%29
3x + 1 = 3.744
3x = 3.744 - 1
3x = 2.744
x = 2.744%2F3
x = .9147
Check this on your calc: enter 2^(3(.9147)+1), results: ~13.4