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The Brainliest Answer!
2log_{10} x-log_{x} (10^{-2})=2(log_{10} x+ \frac{1}{log_{10} x} )
as we know always
A.M \geq G.M
 \frac{log_{10} x+ \frac{1}{log_{10} x}}{2} \geq (log_{10} x \frac{1}{log_{10} x})^{0.5}
{log_{10} x+ \frac{1}{log_{10} x}} \geq2
so the minimum value is 4
2 5 2
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