I am trying to rework Y = X(1 + k)^n and be able to solve for n.
I used to know how.
Many years ago.
Grrrr.
I come up with log(1 + k) no matter what. My recollection is that I can not reduce this further.
Is this correct?
Please note, this is not for homework help. I can’t believe I am stumped by this.
Any good tutorial sites would be appreciated. I’m gonna lose sleep until I figure this out. Damn you brain!
You just need two log properties:
log(x^n) = n*log(x)
log(xy) = log(x) + log(y)
So expanding using that we get, for any logarithm basis: log(y) = log(x) + n*log(1+k)
So (1+k)^n = (k^n) * k.
Since log_b x = n is equivalent to b^n = x, we have log_k (Y/XK) = n.
I’m not sure you can simplify more than that.
Edit: This assumes you mean X(1 + k)^n as written and not [X(1 + k)]^n.
And of course solving this linear equation in n gives:
n = [log(y) - log(x)]/log(1+k) = log_(1+k)(y/x)
Reading that out loud it says that n is the number that, raised over (1+k), equals y/x. Assuming, like the poster above said, that you meant y = x*[(1+k)^n].
But this is incorrect:
k*(k^n) = k^(n+1) != (1+k)^n
Maybe a more direct way to do it would be using the property:
log_a(a^b) = b
So taking the logarithm base (1+k) to both sides of y/x = (1+k)^n would also yield the same answer. But in general calculators don’t have arbitrary log basis so this property comes handy too:
log_a(b) = log(b)/log(a) for any log basis on the right side.