Question

$\mathbf{\sum }{\mathbf{lnA}}_{\mathbf{k}}\mathbf{=}\mathbf{ln}\prod A_k$

(a) True

(b)False

Short answer

Therefore, the correct option is (b) True.

Step-by-step solution

Step 1:  Check for left hand side  ∑lnAk

Let $\mathbf{k}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{.}\mathbf{..}\mathbf{,}\mathbf{N}$

Therefore$\sum _{\mathrm{k}=1}^{\mathrm{N}}{\mathrm{lnA}}_{\mathrm{k}}$is,

$\begin{array}{c}\sum _{k=1}^N{\mathbf{lnA}}_k\mathbf{=}{\mathbf{lnA}}_1+{\mathbf{lnA}}_2+{\mathbf{lnA}}_3+....+{\mathbf{lnA}}_N\\ \mathbf{=}\mathbf{ln}({\mathbf{A}}_1{\mathbf{A}}_2{\mathbf{A}}_3....{\mathbf{A}}_N)\end{array}$…… (1)

Check for right hand side  ln∏Ak

Now,$\ln \prod _{\mathrm{k}=1}^{\mathrm{N}}{\mathrm{A}}_{\mathrm{k}}$is,

$\ln \prod _{\mathrm{k}=1}^{\mathrm{N}}{\mathrm{A}}_{\mathrm{k}}=\ln ({\mathrm{A}}_1{\mathrm{A}}_2{\mathrm{A}}_3+...{\mathrm{A}}_{\mathrm{N}})$…… (2)

Therefore, from equations (1) and (2)$\sum \ln A_k=\ln \prod _{}{A}_k$

Hence, correct option is (b)True.