Evaluate the following:
2 + 12\dfrac{1}{2}21log (10)-3
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Given,
⇒2+12log(10)−3⇒2+−32log10⇒2−32×1⇒2−32=12.\Rightarrow 2 + \dfrac{1}{2}\text{log} (10)^{-3} \\[1em] \Rightarrow 2 + \dfrac{-3}{2}\text{log}10 \\[1em] \Rightarrow 2 - \dfrac{3}{2} \times 1 \\[1em] \Rightarrow 2 - \dfrac{3}{2} = \dfrac{1}{2}.⇒2+21log(10)−3⇒2+2−3log10⇒2−23×1⇒2−23=21.
Hence, 2+12log(10)−3=122 + \dfrac{1}{2}\text{log} (10)^{-3} = \dfrac{1}{2}2+21log(10)−3=21.
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