The solution an algebraic equation.?

It would be useful to memorize the following:

lim (1 + 1/x)^x = e

x→∞ 

let x = n/r

(1+r/n)^(nt) = (1 + 1/x)^(xrt)

= ((1 + 1/x)^(x))^(rt)

lim ((1 + 1/x)^(x))^(rt) = e^(rt)

x→∞

Thus

2P = Pe^(rt)

t = ln(2)/r

Because ln(2) is approximately 70% (and less-frequent compounding increases t),  this leads to the "Rule of 72": for a given rate of return r%, your investment will double in approximately 70/r or 72/r years.

A=P(1+r/n)^(nt)

=>

ln(A)=ln(P)+ntln(1+r/n)

=>

ln(A/P)=ln(1+r/n)/[1/(nt)]

=>

ln(A/P)->[1/(1+r/n)](-r/n^2)/[(-1/n^2)/t]

as n->infinity

=>

ln(A/P)->rt

=>

A=Pe^(rt)

When A=2P

2P=Pe^(rt)

=>

t=ln(2)/r

[Note that dependence of memorizing

formula is not a goodway because there

are a lot , a lot of formula in math! It  is

a good habit to derive formula as much

as possible when needed. Of course,

some basic formulas are needed to

remember as the basis]

I'm not sure what you are asking.

If you are saying to solve for t from:

A = P(1 + r/n)^(nt)

and A = 2P, then we can do the substitution first, then divide both sides by P:

2P = P(1 + r/n)^(nt)

2 = (1 + r/n)^(nt)

Now get the log of both sides:

ln(2) = ln[(1 + r/n)^(nt)]

we can pull the exponent out of the log:

ln(2) = nt ln(1 + r/n)

And divide both sides by [n ln(1 + r/n)]:

ln(2) / [n ln(1 + r/n)] = t

 A=P(1+r/n)^(nt)

2P = P(1 + r/n)^(nt)

(1 + r/n)^(nt) = 2P/P

(1 + r/n)^(nt) = 2

ntln(1 + r/n) = ln(2)

...........ln(2)

t = ------------------

........nln(1 + r/n)

........... ln(2)

t = ----------------------- Answer//

.......... nln[(n + r)/n]