Question

How do you evaluate $1200=300{{\left( 1+r \right)}^{5}}$?

Answer 1

Hint: We begin by dividing both sides of the equation by 300. We then raise the exponent of both sides to $\dfrac{1}{5}$. We then use law exponent involving power raised to power ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ to simplify both sides. We simplify to ensure $r$ remains at only one side of the equation.

Complete step by step solution:
We know from the law of exponent of power raised to a power for the same base that the powers are multiplied with the same base. If $a$ is the ${{\left( {{a}^{m}} \right)}^{n}}$ is power $m$ raised to the power $n$ then by this law we have;
\[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\]
We are given in the following equation in the question
\[1200=300{{\left( 1+r \right)}^{5}}\]
We divide both sides of the above equation by 300 to have;
\[\begin{align}
  & 4={{\left( 1+r \right)}^{5}} \\
 & \Rightarrow {{2}^{2}}={{\left( 1+r \right)}^{5}} \\
\end{align}\]
Let us raise both sides of the equation to the power $\dfrac{1}{5}$ to have
\[\Rightarrow {{\left( {{2}^{2}} \right)}^{\dfrac{1}{5}}}={{\left( {{\left( 1+r \right)}^{5}} \right)}^{\dfrac{1}{5}}}\]
We use the law of exponent of power raised to power ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ for $a=2,m=2,n=\dfrac{1}{5}$ in the left hand side and for $a=1+r,m=5,n=\dfrac{1}{5}$ in the right hand side of the above step to have;
\[\begin{align}
  & \Rightarrow {{2}^{2\times \dfrac{1}{5}}}={{\left( 1+r \right)}^{5\times }}^{\dfrac{1}{5}} \\
 & \Rightarrow {{2}^{\dfrac{2}{5}}}=1+r \\
 & \Rightarrow 1+r={{2}^{\dfrac{2}{5}}} \\
\end{align}\]
We subtract both sides of above equation by 1 to have;
\[\Rightarrow r={{2}^{\dfrac{2}{5}}}-1\]
The above value is the evaluated with a calculator to have
\[r\simeq 1.32-1=0.32\]

Note: We can alternatively evaluate using logarithm. We divide both sides of the give equation just like above to have;
\[\begin{align}
  & 4={{\left( 1+r \right)}^{5}} \\
 & \Rightarrow {{2}^{2}}={{\left( 1+r \right)}^{5}} \\
\end{align}\]
We take common logarithm both sides of the above equation to have;
\[\Rightarrow \log {{2}^{2}}=\log {{\left( 1+r \right)}^{5}}\]
We use the logarithmic identity involving power $\log {{x}^{m}}=m\log x$ for $x=2,m=2$ in the left hand side and for $x=1+r,m=5$in the above step to have;
\[\begin{align}
  & \Rightarrow 2\log 2=5\log \left( 1+r \right) \\
 & \Rightarrow \log \left( 1+r \right)=\dfrac{2\log 2}{5} \\
\end{align}\]
We us the value of $\log 2=0.3010$ in the above step to have
\[\begin{align}
  & \Rightarrow \log \left( 1+r \right)=\dfrac{2\times .301}{5} \\
 & \Rightarrow \log \left( 1+r \right)=.1204 \\
\end{align}\]
We take ant-logarithm both sides of the above step to have
\[\begin{align}
  & \operatorname{antilog}\left( \log \left( 1+r \right) \right)=\operatorname{antilog}\left( 0.1204 \right) \\
 & \Rightarrow 1+r\simeq 1.32 \\
 & \Rightarrow r\simeq 0.32 \\
\end{align}\]
We require antilog table to evaluate above value.