Question

If x>1 the least value of 2log base 10 x-logBase x 0.01 is​

Answer 1

Question: If x>1, what is the least value of $2log_{10}x-log_{x}0.01$?

Let's convert the base using the log rule.

$2log_{10}x-log_{x}0.01$

$=2\dfrac{log_{10}x}{log_{10}10} -\dfrac{log_{10}0.01}{log_{10}x}$

Replace $t$ for $log_{10}x$.

$\dfrac{2t}{log_{10}10} -\dfrac{log_{10}0.01}{t}$

$=2t-\dfrac{-2}{t}$

$=2t+\dfrac{4}{t}$

Since $x>1$, the value of $log_{10}x$ is positive.

So, $2t>0$ and $\dfrac{4}{t} >0$.

Let's find the minimum value using the A.M-G.M inequality.

$\bigstar \dfrac{a+b}{2} \geq\sqrt{ab}\:\:\mathrm{(A.M-G.M\:Inequality)}$

$\implies\dfrac{1}{2} (2t+\dfrac{4}{t} )\geq \sqrt{2t\times\dfrac{4}{t} }$

$\implies 2t+\dfrac{4}{t} \geq 4\sqrt{2}\:\:\mathrm{(Equals\:where\:t=\sqrt{2} )}$

Returning to the original question, the maximum value of $2log_{10}x-log_{x}0.01$ is $4\sqrt{2}$, where $x=10^{\sqrt{2}}$.

More information:

The A.M-G.M inequality applies to two positive numbers.

The expression equals where two numbers are equal.

This fact is used in these types of problems:

• It can be used when the product is constant.

• It can be used when you're given the addition.

We can use it to find the maximum area of the quadrilateral, or we can use it to find the minimum value of $2t+\dfrac{4}{t} \:(t>0)$.