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

User Yetunde
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1 Answer

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Answer:


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

User Pavel Synek
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