Question

If ${5^{56}}{\left( {\dfrac{1}{5}} \right)^{\text{x}}}{\left( {\dfrac{1}{5}} \right)^{\sqrt {\text{x}} }}$> 1, then x belongs toA. (0, 49)B. (49, 64)C. (0, 64)D. (49, 64)

Hint: Any number to the power of zero is equal to one. The square root of any rational number is always greater than or equal to zero. If the multiple of two numbers is less than zero then only one of the two numbers could be less than zero.

Given Data –
${5^{56}}{\left( {\dfrac{1}{5}} \right)^{\text{x}}}{\left( {\dfrac{1}{5}} \right)^{\sqrt {\text{x}} }}$> 1,

The given inequality is only true for x≥0.

The laws of exponents states that for a given number ‘a’
${\text{1}}{\text{. }}{{\text{a}}^{\text{m}}} \times {{\text{a}}^{\text{n}}} = {{\text{a}}^{\left( {{\text{m + n}}} \right)}} \\ 2.\dfrac{{{{\text{a}}^{\text{m}}}}}{{{{\text{a}}^{\text{n}}}}} = {{\text{a}}^{\left( {{\text{m - n}}} \right)}} \\$
$\Rightarrow {\left( {\dfrac{1}{5}} \right)^{\text{x}}}{\left( {\dfrac{1}{5}} \right)^{\sqrt {\text{x}} }} = \left( {\dfrac{{{1^{\text{x}}} \times {1^{\sqrt {\text{x}} }}}}{{{5^{\text{x}}} \times {5^{\sqrt {\text{x}} }}}}} \right) \\ \\$

One to the power of anything equals to one itself.
$\Rightarrow \left( {\dfrac{1}{{{5^{{\text{x + }}\sqrt {\text{x}} }}}}} \right)$

Now our inequality looks like this,
${5^{56}}\left( {\dfrac{1}{{{5^{{\text{x + }}\sqrt {\text{x}} }}}}} \right) > 1 \\ \Rightarrow {5^{56{\text{ - x - }}\sqrt {\text{x}} }} > 1 \\$
1 can be expressed as ${5^0}$ (5 to the power of zero)
$\Rightarrow {5^{56 - {\text{x - }}\sqrt {\text{x}} }} > {5^0} \\ \Rightarrow 56 - {\text{x - }}\sqrt {\text{x}} > 0 \\ {\text{ }} \\$
$56 - {\text{x - }}\sqrt {\text{x}}$Can be expressed as ${\text{56 - x + 7}}\sqrt {\text{x}} {\text{ - 8}}\sqrt {\text{x}}$
$\Rightarrow 7\left( {\sqrt {\text{x}} + 8} \right) - \sqrt {\text{x}} \left( {\sqrt {\text{x}} + 8} \right) \\ \Rightarrow \left( {\sqrt {\text{x}} + 8} \right)\left( {7 - \sqrt {\text{x}} } \right) > 0 \\ \Rightarrow \left( {\sqrt {\text{x}} + 8} \right)\left( {\sqrt {\text{x}} - 7} \right) < 0 \\$

As $\left( {\sqrt {\text{x}} + 8} \right)$is positive for all rational numbers, that means
$\left( {\sqrt {\text{x}} - 7} \right)$Should be negative. (i.e. <0)
$\Rightarrow \sqrt {\text{x}} < 7$
Squaring on both sides
$\Rightarrow {\text{x}} < 49 \\ \Rightarrow 0 \leqslant {\text{x < 49}} \\$

Hence x belongs to (0, 49) which makes Option A the correct answer.

Note –

In such types of questions first simplify the equation by bringing all the exponents of the same number together using the laws of exponents, then compare it as a whole to the other part of the inequality. Then simplify the equation accordingly and find out the factors if there are any in order to obtain the limits of the required number. Properties of rational numbers come in handy while solving these types of questions.