Q. 78 - Calculus

Question

Prove $e^{a+b}=e^a+e^b \mathrm{for} \mathrm{any} \mathrm{real} \mathrm{number} \mathrm{a},\mathrm{b}$

Step-by-Step Solution

Answer

$e^{a+b}=e^a+e^b \mathrm{for} \mathrm{any} \mathrm{real} \mathrm{number} \mathrm{a},\mathrm{b}$ Proved

Step 1. To prove

$e^{a+b}=e^a+e^b \mathrm{for} \mathrm{any} \mathrm{real} \mathrm{number} \mathrm{a},\mathrm{b}$.

Step 2. Solving LHS

$$\begin{gathered} Let,y=e^{a+b} \\ Then, \\ y=e^{a+b} \\ \mathrm{Iny}=\ln \left(e^{a+b}\right) [\mathrm{Taking} \mathrm{In} \mathrm{on} \mathrm{both} \mathrm{sides}] \\ \mathrm{In}y=a+b ln\left(e^1\right)=1 \\ \mathrm{Therefore},\mathrm{Iny}=a+b. \end{gathered}$$

Step 3. RHS

$$\begin{gathered} Let,z=e^a{e}^t \\ \mathrm{Then},z=e^a{e}^b \\ ln\left(z\right)=ln\left(e^a{e}^b\right) [\mathrm{Taking} \mathrm{In} \mathrm{on} \mathrm{both} \mathrm{sides}]\mathrm{l} \\ \ln \left(z\right)=ln\left(e^a\right)+\ln e^{\mathrm{b}}[\mathrm{property} \mathrm{of} \mathrm{logarithm}] \\ ln\left(z\right)=a+b \left[ln\right(e^1)=1] \\ \mathrm{Therefore},\mathrm{In}z=a+b. \end{gathered}$$

Step 4. Result

$$\begin{gathered} \mathrm{The} \mathrm{results} \mathrm{of} \mathrm{part} \mathrm{a} \mathrm{and} \mathrm{part} \mathrm{b} \mathrm{are}. \\ ln\left(y\right)=a+b \\ ln\left(z\right)=a+b \\ So,lnz=lny y=z \\ \mathrm{Therefore},e^{a+b}=e^a{e}^b \end{gathered}$$