Question

In the equation for the exponential function, f ( x ) = P ( 1 + r ) ^ x, how do we find r (rate of change) algebraically?

Answer 1

`r=\left((y)/(P)\right)^{(1)/(x)}-1`

Explanation:

Given exponential function:

`f(x)=P(1+r)^x`

Replace f(x) with y:

`y=P(1+r)^x`

Divide both sides by P:

`(y)/(P)=(1+r)^x`

Take natural logs of both sides:

`\ln \left((y)/(P)\right)=\ln (1+r)^x`

`\textsf{Apply the power law}: \quad \ln x^n=n \ln x`

`\ln \left((y)/(P)\right)=x \ln (1+r)`

Divide both sides by x:

`(1)/(x)\ln \left((y)/(P)\right)=\ln (1+r)`

`\textsf{Apply the power law}: \quad n \ln x=\ln x^n`

`\ln \left((y)/(P)\right)^{(1)/(x)}=\ln (1+r)`

Raise both sides to power of e:

`\large\text{$e^{\ln \left((y)/(P)\right)^{(1)/(x)}}=e^(\ln (1+r))$}`

`\textsf{Apply the power law}: \quad e^(\ln x)=x`

`\left((y)/(P)\right)^{(1)/(x)}=1+r`

Subtract 1 from both sides:

`\left((y)/(P)\right)^{(1)/(x)}-1=r`

Therefore, the equation to find r is:

`r=\left((y)/(P)\right)^{(1)/(x)}-1`