Question

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

Short answer

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

Step-by-step solution

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

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

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

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