Ask Dr. LJ

[evelio]

I can't remember High School Math enough

50 = T * (1.01)^T

Is there a way to solve for T without graphing?

Actually what I'm trying to figure out is:

V = C * T * (1+R)^T

I know V, C and R, and want to calculate T.

Anybody?

Comments

[faerieloch]

Um. There's something about raising everything to an exponent of e:
e^50=e^(T*1.01^T)
which equals
e^50=T*e^(T*1.01)

In all honesty, though, I'm guessing. And I can't remember the inverse. :( I think it's ln....

[jadia]

log(V) = log(CT)+T*log(1+R), I believe.

So: T = [log(V) - log(CT)]/(log(1+R)).

log can be in any base, e or 10 or whatever.

[jadia]

Basically:

log(A*B) = log(A)+log(B)
log(A^B) = B*log(A)

[evelio]

But T still depends on log(CT). So I can't really solve for it... unless I'm missing something.

[jadia]

oops, you're right.

Hmmm.

[quaintance]

I got my staff of applied and theoretical physicists on it. You can solve by making the assumption that (1 + R)^T = 1 + RT for values of R << 1.

So...

50 = T(1 + 0.01T)
50 = T + 0.01T^2
0 = 0.01T^2 + T - 50

Then use the quadratic...
so...

T = 36.6