Evaluate the following without using log tables :

5 log 2 + $\dfrac{3}{2}$ log 25 + $\dfrac{1}{2}$ log 49 − log 28

Given,

⇒ 5 log 2 + $\dfrac{3}{2}$ log 25 + $\dfrac{1}{2}$ log 49 − log 28

$\Rightarrow 5 \log \space 2 + \dfrac{3}{2} \log \space 25 + \dfrac{1}{2} \log \space 49 − \log \space 28 \\[1em] \Rightarrow \log \space 2^5 + \log \space 25^\dfrac{3}{2} + \log \space 49^\dfrac{1}{2} − \log \space 28 \\[1em] \Rightarrow \log \space 32 + \log \space (5^2)^{\dfrac{3}{2}} + \log \space \sqrt{49} − \log \space 28 \\[1em] \Rightarrow \log \space 32 + \log \space 5^3 + \log \space 7 − \log \space 28 \\[1em] \Rightarrow \log \space 32 + \log \space 125 + \log \space 7 − \log \space 28 \\[1em] \Rightarrow \log \space (32 × 125 × 7) − \log \space 28 \\[1em] \Rightarrow \log \space 28000 − \log \space 28 \\[1em] \Rightarrow \log \space \Big(\dfrac{28000}{28}\Big) \\[1em] \Rightarrow \log \space 1000 \\[1em] \Rightarrow \log \space 10^3 \\[1em] \Rightarrow 3 \log \space 10 \\[1em] \Rightarrow 3 \times 1 \\[1em] \Rightarrow 3.$

Hence, 5 log 2 + $\dfrac{3}{2}$ log 25 + $\dfrac{1}{2}$ log 49 − log 28 = 3.