Question

Let the domain of ‘x’ for the given inequation, ${{\log }_{5x+4}}\left( {{x}^{2}} \right)\le {{\log }_{5x+4}}\left( 2x+3 \right)$ be (m, n). Find m+n?

Answer 1

Hint:Apply logic and find simple inequalities for which x satisfies such as $5x+4>1,x\ge 0$and ${{x}^{2}}\le 2x+3$, from each of them find the values of x and then get common value and hence get the result.

Complete step-by-step answer:
In the question we are given that,
${{\log }_{\left( 5x+4 \right)}}{{x}^{2}}\le {{\log }_{\left( 5x+4 \right)}}\left( 2x+3 \right)$
For the above equation to be logically possible and first condition for x will be,
5x + 4 > 1
Otherwise if the 5x + 4 lies between 0 and 1 then a negative sign will come in the equation and the equation may not be valid.
So by inequation, we have
5x + 4 > 1
5x > 1-4
So, 5x > -3
Hence, $x>\dfrac{-3}{5}...........(i)$
Now we are using our 2nd condition which is ${{x}^{2}}>0$, we get
 x > 0………(ii)
Now, we will use the rule to find third condition which is,
If ${{\log }_{a}}b\ge {{\log }_{a}}c$ and both ${{\log }_{a}}b$ and ${{\log }_{a}}c$ is greater than 0 then we can say that,
$b\ge c.$
So, by using it we get,
${{x}^{2}}\le 2x+3$
Now by taking (2x + 3) to other side of in equation we get,
${{x}^{2}}-2x-3\le 0$
By using middle term factor we can write it as,
${{x}^{2}}-3x+x-3\le 0$
By further factoring we get,
$\left( x+1 \right)\left( x-3 \right)\le 0$
Therefore, for the inequality to be true, x should lie in the range [-1, 3]
Now from the three conditions so obtained above, we got the value as
$x>\dfrac{-3}{5},x>0$and$x\in \left[ -1,+3 \right]$
Now, by taking common values of ‘x’, we get ‘x’ as ‘(0, 3)’, this means x will be greater than 0 and less than or equal to 3.
In the question range of x was given as (m, n).
So, we can now tell that m is ‘0’ and n is ‘3’
Hence, the value of m + n is (0 + 3) = 3

Note: While using or writing inequalities please imagine it and think logically whether it is possible or not, then apply it. The inequalities related to logarithms are generally tricky so students have to carefully avoid any mistake.
Another approach is, in the given equation the base of both the ‘log’ is same so we can write,
${{x}^{2}}\le \left( 2x+3 \right)$
This way also we will get the same result.