一條中五數學

The equation is:

$20000 x (0.64)^t - $11000 x 0.8^t = $3000
$20 x (0.64)^t - $11 x 0.8^t = $3
Solve for t.

Do by trial and error
When t = 1, $20 x (0.64)^1 - $11 x 0.8^1 = 12.8 – 8.8 = 4
When t = 2, $20 x (0.64)^2 - $11 x 0.8^2 = 8.192 – 7.04 = 1.152
T lies between 1 and 2
When t = 1.5, $20 x (0.64)^1.5 - $11 x 0.8^1.5 = 10.24 – 7.87 = 2.37
When t = 1.29, $20 x (0.64)^1.29 - $11 x 0.8^1.29 = 11.25 – 8.25 = 3.00

Check:
$20000 x (0.64)^1.29 - $11000 x 0.8^1.29 = $11246.10 – 8248.57 = 2997.53

When t = 1.289
$20000 x (0.64)^1.289 - $11000 x 0.8^1.289 = $11251.12 – 8250.41 = 3000.71
When t = 1.289225
$20000 x (0.64)^ 1.289225 - $11000 x 0.8^1.289225 = $11250.00 – 8250.00 = 3000.00

t 是 1.289225年 (1年3個月14 日)

If you know Calculus, use Newton’s Method
Let the first approximation be t = 1.29
y = 20000 x (0.64)^t - $11000 x 0.8^t - 3000
dy/dt = 20000 x (0.64)^t ln0.64 - $11000 x 0.8^t ln 0.8
dy/dt = (11246.10) ln0.64 - (8248.57) ln 0.8 = -5018.989 +1840.61 = -3178.37
Xn+1 = Xn – f(Xn)/f’(Xn)
Xn+1 = 1.29 – (-2.47)/ (-3178.37) = 1.289222

Comment:
Alternatively, you can use spreadsheet Excel or write a computer program to solve it.