Question

For any base, show that $\log \left(1\times 2\times 3\right)=\log \left(1\right)+\log \left(2\right)+\log \left(3\right)$

Answer 1

Step 1: Solve $\mathrm{LHS}$

$\mathrm{LHS}=\log (1\times 2\times 3)$

$=\log \left(6\right)$ …$\left[1\right]$

Step 2: Solve $\mathrm{RHS}$

$\mathrm{RHS}=\log \left(1\right)+\log \left(2\right)+\log \left(3\right)$

We know that $\mathbf{log}\mathbf{\left(}\mathit{m}\mathbf{\right)}\mathbf{+}\mathbf{log}\mathbf{\left(}\mathit{n}\mathbf{\right)}\mathbf{=}\mathbf{log}\mathbf{(}\mathit{m}\mathbf{\times }\mathit{n}\mathbf{)}$

$\begin{array}{rcl}\therefore \log \left(1\right)+\log \left(2\right)+\log \left(3\right)& =& \log \left(1\times 2\times 3\right)\end{array}$

$\begin{array}{rcl}& =& \log \left(6\right)\end{array}$ ….$\left[2\right]$

From $\left[1\right]$ and $\left[2\right]$,

$\mathrm{LHS}=\mathrm{RHS}$

Hence proved that $\log (1\times 2\times 3)=\log \left(1\right)+\log \left(2\right)+\log \left(3\right)$