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Question

# For any base, show that $\mathrm{log}\left(1×2×3\right)=\mathrm{log}\left(1\right)+\mathrm{log}\left(2\right)+\mathrm{log}\left(3\right)$

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Solution

## Step 1: Solve $\mathrm{LHS}$$\mathrm{LHS}=\mathrm{log}\left(1×2×3\right)$ $=\mathrm{log}\left(6\right)$ …$\left[1\right]$Step 2: Solve $\mathrm{RHS}$$\mathrm{RHS}=\mathrm{log}\left(1\right)+\mathrm{log}\left(2\right)+\mathrm{log}\left(3\right)$We know that $\mathbf{log}\mathbf{\left(}\mathbit{m}\mathbf{\right)}\mathbf{+}\mathbf{log}\mathbf{\left(}\mathbit{n}\mathbf{\right)}\mathbf{=}\mathbf{log}\mathbf{\left(}\mathbit{m}\mathbf{×}\mathbit{n}\mathbf{\right)}$$\begin{array}{rcl}\therefore \mathrm{log}\left(1\right)+\mathrm{log}\left(2\right)+\mathrm{log}\left(3\right)& =& \mathrm{log}\left(1×2×3\right)\end{array}$ $\begin{array}{rcl}& =& \mathrm{log}\left(6\right)\end{array}$ ….$\left[2\right]$ From $\left[1\right]$ and $\left[2\right]$,$\mathrm{LHS}=\mathrm{RHS}$Hence proved that $\mathrm{log}\left(1×2×3\right)=\mathrm{log}\left(1\right)+\mathrm{log}\left(2\right)+\mathrm{log}\left(3\right)$

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