Answers

The Brainliest Answer!
2014-04-26T11:30:57+05:30
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