Question

Solve ${{\log }_{0.2}}\left| x-3 \right|\ge 0$ . What is the maximum possible value that x can take?

Answer 1

- Hint: Use general logarithmic properties to solve this logarithm inequality. First check the basic conditions for a logarithm if it exists. If it satisfies the basic necessary condition, then we can say its value exists. Now, send the $\log$ term on the left hand side of inequality to make it into the base of a power. Now use general algebraic techniques to solve the inequality. After solving by removing the modulus, you will get a range of values possible to x. Now, pick the maximum value in that range. This value is the required result.

Complete step-by-step solution -

For a logarithm to exist we need two basic conditions:
${{\log }_{a}}b$ to exist $a>0$ ……………………………….(1)
${{\log }_{a}}b$ to exist $b>0$ …………………………………(2)
${{\log }_{a}}b$ to exist $a\ne 1$
Some properties of logarithm are given by formulae given below:
${{a}^{{{\log }_{a}}b}}=b$ …………………………………..(3)
${{\log }_{a}}b\ge k$
Take a raise to power on both sides, you get inequality:
${{a}^{{{\log }_{a}}b}}\ge {{a}^{k}}$
By substituting equation (3) in this equation, we get:
$b\ge {{a}^{k}}$ …………………………………(4)
Given expression in the question which we need to solve is:
$\Rightarrow {{\log }_{0.2}}\left| x-3 \right|\ge 0$ ………………………………(5)
First check the condition (1), is it satisfy or not:
$a=0.2,$ $a>0$
Next check condition (2), make it satisfy for logarithm to exist:
$\Rightarrow \left| x-3 \right|>0$
By simplifying the above inequality, we can say that:
$x\ne 3$
By taking the left-hand side term and right-hand side term to the power of 0.2, we get:
$\Rightarrow {{0.2}^{{{\log }_{0.2}}\left| x-3 \right|}}\ge {{0.2}^{0}}$
By using equation (3) in the above equation, we get the inequality:
$\Rightarrow \left| x-3 \right|\ge 1$
By general modulus property, we can write the inequalities:
$x-3\ge 1$ $-\left( x-3 \right)\le 1$
By simplifying these inequalities, we get these inequalities of x:
$x\ge 4$ $x\le 2$
By above we can write range of x, as follows:
$\Rightarrow x\in \left( -\infty ,2 \right]\cup \left[ 4,\infty \right)$
Therefore, this range is possible for values of x.

Note: Be careful while applying the properties of logarithm, because generally students forget to apply condition $a,b>0$. By this you may get extra roots which will satisfy inequality but at that place logarithm value does not exist. Here, the value $x\ne 3$ does not lie in the range so no problem but if it is in range then you must exclude that term.