Five thousand dollars compounded annually at an x% interest rate takes six years to double. at the same interest rate, how many years will it take $300 to grow to $9600?

$5,000,00 deposited for 6 years becomes $10,000.00.
What is the interest rate?

The formula you need requires logarithms:
log(1 + rate) = {log(total) -log(Principal)} ÷ Years
log(1 + rate) = {log(10,000) -log(5,000)} ÷ 6
log(1 + rate) = (4 -3.6989700043 ) / 6
log(1 + rate) = 0.301029957 / 6
log(1 + rate) = 0.0501716595
Now we get the number 10 and raise it to 0.0501716595 which equals
1.1224620317
which is 1 plus the rate so the rate equals
.1224620317 or 12.24620317 per cent
That's the end of Part ONE.
Now, on to Part TWO.

How many years does it take for 300 to become 9,600 at an annual rate of 12.24620317%?
You will need to use this formula:
(Yes, more logarithms).
Years = {log(total) -log(Principal)} ÷ log(1 + rate)
Years = {log(9,600) - log(300)} / log(1.1224620317)
Years = (3.982271233 -2.4771212547) / 0.050171659518
Years = 1.5051499783 /  .050171659518
Years = 30

y-int:  -6

x-int: -2/5 and 3

Step-by-step explanation:

you can find the y-intercept by substituting zero in for every 'x' in the equation;

y = 5(0) - 13(0) - 6

y = -6

to find the x-intercept(s), factor the trinomial by trying to find a pair of factors that multiply to 6 and add to -13; the factors that work are:

(5x + 2)(x - 3)

now set each factor equal to zero and solve for 'x':

5x + 2 = 0

5x = -2

x = -2/5

x - 3 = 0

x = 3