If {log _{10}}x = y,then {log _{1000}}{x^2} i | vedclass

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(D) Given that ${\log _{10}}x = y$. We need to evaluate ${\log _{1000}}{x^2}$. Using the change of base formula and properties of logarithms: ${\log _

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$\frac{2y}{3}$

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(D) Given that ${\log _{10}}x = y$. We need to evaluate ${\log _{1000}}{x^2}$. Using the change of base formula and properties of logarithms: ${\log _{1000}}{x^2} = \frac{{\log _{10}}{x^2}}{{\log _{10}}{1000}}$ $= \frac{2{\log _{10}}x}{{\log _{10}}{{10}^3}}$ $= \frac{2{\log _{10}}x}{3{\log _{10}}10}$ Since ${\log _{10}}10 = 1$,we have: $= \frac{2}{3}{\log _{10}}x = \frac{2}{3}y$.

If ${\log _{10}}x = y$,then ${\log _{1000}}{x^2}$ is equal to

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