Can anyone transpose this equation to solve for n as a function of A, P, i ie, n = .....
A = P(((1 + i)^{n }- 1)/( (1 - I) + (1 + i)^{n}))
Please show each step!
Sorry for the typo! Added = sign after A
Can anyone transpose this equation to solve for n as a function of A, P, i ie, n = .....
A = P(((1 + i)^{n }- 1)/( (1 - I) + (1 + i)^{n}))
Please show each step!
Sorry for the typo! Added = sign after A
$A = P\dfrac{(1+i)^n-1}{(1-i)+(1+i)^n}$
$\dfrac{A}{P} = \dfrac{(1-i)-(1-i)+(1+i)^n-1}{(1-i)+(1+i)^n} = 1+\dfrac{i-2}{(1-i)+(1+i)^n}$
$\dfrac{A}{P}-1 = \dfrac{i-2}{(1-i)+(1+i)^n}$
$\left(\dfrac{A}{P}-1\right) ((1-i)+(1+i)^n) = i-2$
$\dfrac{A(1-i)}{P}-1+i+\left(\dfrac{A}{P}-1\right) (1+i)^n = i-2$
$\left(\dfrac{A}{P}-1\right) (1+i)^n = \dfrac{A(i-1)}{P}-1$
$(1+i)^n = \dfrac{\dfrac{A(i-1)}{P}-1}{\dfrac{A}{P}-1}$
$n = \dfrac{ \ln \left( \dfrac{A(i-1)}{P} - 1\right) - \ln \left( \dfrac{A}{P} - 1 \right) }{\ln (1+i) }$
Those commands have nothing to do with mathematics.
They are type-settings to make the symbols.
1)n = \dfrac{ \ln \left( \dfrac{A(i-1)}{P} - 1\right) - \ln \left( \dfrac{A}{P} - 1 \right) }{\ln (1+i) } is the code required to see
2)$n = \dfrac{ \ln \left( \dfrac{A(i-1)}{P} - 1\right) - \ln \left( \dfrac{A}{P} - 1 \right) }{\ln (1+i) }$
Line #1) is exactly the same as line #2) except 2) begin with and ends with $. That tell the browser to display LaTex.
SEE HERE
You probably can't. As far as I have been able to figure out, Word does not support LaTeX but has its own (not very good in my opinion) mathematical option.
Do you see a summation symbol when you look below?
$\displaystyle \sum_{j=1}^nx^j$
If so, then your browser is rendering LaTeX properly.
If instead you see \displaystyle \sum_{j=1}^nx^j, your browser is not rendering LaTeX properly.