CPhill

Thanks for that answer, Bertie.....I went through the algebra as you suggested.......that's an interesting result to note that the product of both lines gives us the rotated conic equation.....I suspect that this might just be a coincidence, albeit a pretty fascinating one ....

Two questions......

The normal rotated conic usually contains an x and a y term, but I noticed that you omit them in your assumed equation........why is this???......

Also.....

Why do we let X and Y approach infinity??  ....Is this just to eliminate the c/X^2  term so we can solve the quadratic that follows??  I have seen Alan use this technique in solving another hyperbola problem......I was never able to wrap my head around it.....!!!!

All in all....this is a  really neat problem......

BTW......graphing this in Desmos just produces a pair of intersecting lines  .....https://www.desmos.com/calculator/quxgucllfk     ....????

OK....assuming end of year deposits, we have

1(1.05)^99 + 2(1.05)^98  + 3(1.05)^97 +  ....+  99(1.05)^1 + 100(1.05)^0

n = 100

∑      (n)(1.05)^(100-n)  ≈  $52,810.53

n = 1

Yeah, Melody....I see that I'm probably wrong.....I'm just trying to figure out my mistake....do you see it???   ....did I "double-count" somehow???

[ My answer seems  "logical"......LOL!!!! ]

Thanks, Melody....!!!

Assuming that the deposits are made at the beginning of each year ....we have....

1(1.05)^100  + 2(1.05)^99  + 3(1.05)^98  +   ...... +  99(1.05)^2  + 100(1.05)

n = 99

∑         (n + 1)(1.05)^(100-n)       =  about $55,451.05

n = 0

Maybe heureka knows a shortcut ???

I believe this is supposed to  be  x √(x+(√x-2))=2

x √(x+(√x-2))= 2   divide both sides by x

√(x+(√x-2))  = 2/x  squre both sides

x + √x  - 2  =  4/x^2

√x  = 4/x^2  - x  + 2   

√x  = [ 4 - x^3  + 2x^2] / x^2      multiply both sides by x^2

x^(5/2)  = 4 - x^3 + 2x^2

x^3 + x^(5/2) -2x^2 - 4  = 0     

WolframAlpha shows the solution to this as   x ≈ 1.83436

log₂24-log₂3=log₅x     using a log rule, we can write

log₂ [ 24/3]  = log₅x

log₂ [8]  = log₅x

3  = log₅x

This says that, in exponential form,

5^3  = x   =  125

OK......but shouldn't that be,  " Thanks a 1 x 10^6  ??  "

Hit the equal sign twice

Example

Multiplying 6.7 *  6.999  and hitting the equal sign twice  gives  4.68933e1  = 4.68933 x 10^1