Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

Let (17)^3.5 × (17)^x = 17⁸
Then, (17)^{3.5 + x} = 17⁸
∴ 3.5 + x = 8
⇒ x = (8 - 3.5)
⇒ x = 4.5

$$\eqalign{ & {\text{Given}}\,{\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{b}{a}} \right)^{x - 3}} \cr & \Rightarrow {\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{a}{b}} \right)^{ - \left( {x - 3} \right)}} = {\left( {\frac{a}{b}} \right)^{\left( {3 - x} \right)}} \cr & \Rightarrow x - 1 = 3 - x \cr & \Rightarrow 2x = 4 \cr & \Rightarrow x = 2 \cr} $$

$$\eqalign{ & {x^z} = {y^2} \Leftrightarrow {10^{\left( {0.48z} \right)}} = {10^{2 \times 0.70}} = {10^{1.40}} \cr & \Rightarrow 0.48z = 1.40 \cr & \Rightarrow z = \frac{{140}}{{48}} = \frac{{35}}{{12}} = 2.9({\text{approx}}) \cr} $$

5^a = 3125     ⇔     5^a = 5⁵
⇒ a = 5.
∴ 5^{(a - 3)} = 5^{(5 - 3)} = 5² = 25

3^{x - y} = 27 = 3³     ⇔     x - y = 3 ....(i)
3^{x + y} = 243 = 3⁵     ⇔     x + y = 5 ....(ii)
On solving (i) and (ii), we get x = 4.

(256)^0.16 × (256)^0.09 = (256)^{(0.16 + 0.09)}
$$\eqalign{ & = {\left( {256} \right)^{0.25}} \cr & = {\left( {256} \right)^{\frac{{25}}{{100}}}} \cr & = {\left( {256} \right)^{\frac{1}{4}}} \cr & = {\left( {{4^4}} \right)^{\frac{1}{4}}} \cr & = {4^1} \cr & = 4 \cr} $$

$$\eqalign{ & {\left( {10} \right)^{150}} \div {\left( {10} \right)^{146}} = \frac{{{{10}^{150}}}}{{{{10}^{146}}}} \cr & = {10^{150 - 146}} \cr & = {10^4} \cr & = 10000 \cr} $$

Given exp. =
$$ = \frac{1}{{\left( {1 + \frac{{{x^b}}}{{{x^a}}} + \frac{{{x^c}}}{{{x^a}}}} \right)}} + $$   $$\frac{1}{{\left( {1 + \frac{{{x^a}}}{{{x^b}}} + \frac{{{x^c}}}{{{x^b}}}} \right)}} + $$   $$\frac{1}{{\left( {1 + \frac{{{x^b}}}{{{x^c}}} + \frac{{{x^a}}}{{{x^c}}}} \right)}}$$
$$ = \frac{{{x^a}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + $$   $$\frac{{{x^b}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + $$   $$\frac{{{x^c}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}}$$
$$\eqalign{ & = \frac{{ {{x^a} + {x^b} + {x^c}} }}{{ {{x^a} + {x^b} + {x^c}} }} \cr & = 1 \cr} $$

$$\eqalign{ & {\text{Let}}\,{\left( {25} \right)^{7.5}} \times {\left( 5 \right)^{2.5}} \div {\left( {125} \right)^{1.5}} = {5^x} \cr & {\text{Then}},\,\frac{{{{\left( {{5^2}} \right)}^{7.5}} \times {{\left( 5 \right)}^{2.5}}}}{{{{\left( {{5^3}} \right)}^{1.5}}}} = {5^x} \cr & \Rightarrow \frac{{{5^{\left( {2 \times 7.5} \right)}} \times {5^{2.5}}}}{{{5^{\left( {3 \times 1.5} \right)}}}} = {5^x} \cr & \Rightarrow \frac{{{5^{15}} \times {5^{2.5}}}}{{{5^{4.5}}}} = {5^x} \cr & \Rightarrow {5^x} = {5^{\left( {15 + 2.5 - 4.5} \right)}} \cr & \Rightarrow {5^x} = {5^{13}} \cr & \therefore x = 13 \cr} $$

$$\eqalign{ & {\left( {0.04} \right)^{ - 1.5}} = {\left( {\frac{4}{{100}}} \right)^{ - 1.5}} \cr & = {\left( {\frac{1}{{25}}} \right)^{ - \left( {3/2} \right)}} \cr & = {\left( {25} \right)^{\left( {3/2} \right)}} \cr & = {\left( {{5^2}} \right)^{\left( {3/2} \right)}} \cr & = {\left( 5 \right)^{2 \times \left( {3/2} \right)}} \cr & = {5^3} \cr & = 125 \cr} $$