1. ## compound interest

hello

please i need help understanding both questions

use the compound interest formula a= p(1+i)^n to answer the following questions:

1 what type of equation is present if n is constant equal to 1, rather than a variable?

2 what type of equation is present if i, A, or P are constant rather than variable?

thanks

2. ## Re: compound interest

Originally Posted by yorkmanz
hello

please i need help understanding both questions

use the compound interest formula a= p(1+i)^n to answer the following questions:

1 what type of equation is present if n is constant equal to 1, rather than a variable?
If $n$ is constant it is a polynomial equation in $p$

2 what type of equation is present if i, A, or P are constant rather than variable?

thanks
There is no $P$ or $A$ in the equation. In $n$ is variable it is exponential in $n$.

3. ## Re: compound interest

for 1 since n is constant i think its just Simple interest and principle

if

n = 1

=A = P(1+i)^n =

=P(1+i)^1

= P(1+i)

now we know that i = r/n

since r n=1,

i = r = interest rate

then we have A = P + P*i

i think its just Simple interest and principle and we have

now the This is a linear equation.

for 2 TRICKY

if i is a constant, this is an exponential equation a=b^n where b=P(1+i) , b a constant. if (In this case, P is a parameter)

and so on for a, b

please correct me if i am wrong

thanks

4. ## Re: compound interest

Originally Posted by yorkmanz
then we have A = P + P*i
One of the points that Walagaster was making is that Mathematics is "case sensitive." "A" and "a" are two different variables. So you need to decide: Is your equation $\displaystyle a = p (1 + i)^n$ or $\displaystyle A = P(1 + i) ^n$. This might seem a minor point but it does need to be addressed.

-Dan