Practice Positive and negative exponent - quantitative aptitude Online Quiz (set-1) For All Competitive Exams

a = 7 - $4√3$

$1/a=1/{7-4√3}$

=$1/{7-4√3}×{7+4√3}/{7+4√3}=7+4√3$

$(√a+1/√a)^2=a+1/a+2$

$=7-4√3+7+4√3+2=16$

$√a+1/√a=4$

$4^61 + 4^62 + 4^63 + 4^64$

= $4^61 (1 + 4 + 4^2 + 4^3)$

= $4^61(1 + 4 + 16 + 64)$

= $4^61$ × 85 which is divisible by 17.

$x = 3^{1/3}- 3^{-1/3}$

On cubing both sides,

$x^3 = ((3)^{1/3})^3 -((3)^{-1/3})^3 -3×3^{1/3}×3^{-1/3}(3^{1/3}- 3^{-1/3})$

$x^3 = 3 - 3^{ - 1}$ - 3x

$x^3 + 3x = 3 - 1/3$

$x^3 + 3x = {9 -1}/3 = 8/3$

$3x^3$ + 9x = 8

$2^{x+4} - 2^{x+2}$ = 3

$2^{x+2}(2^2 - 1)$ = 3

$2^{x+2}$× 3 = 3

$2^{x+2}$ = 1 = 2°

x + 2 = 0 ⇒ x = - 2

$7^1 = 7, 7^2 = 49, 7^3$ = 343,

$7^4 = 2401, 7^5$ = 16807

i.e. after index 4, the unit's digit is repeated.

On dividing 153 by 4, remainder = 1

Unit's digit in the expansion of $(2467)^153=7^1$=7 and unit's digit in the expansion of $(341)^72$ = 1

Required unit's digit = 7 × 1 = 7

$3^{x+8} = 27^{2x+1}$

$3^{x+8} = (3)^{3(2x+1)}$

x + 8 = 6x + 3

5x = 5 ∴ x=1

Unit's digit in the expansion of $(252)^126$

= $2^2$ = 4 (Remainder on dividing 126 by 4 = 2)

Unit's digit in the expansion of $(244)^152$ = 6

Unit's digit in the expansion of $252^126 + 244^152$ = 0

Required remainder = 0

$27^{2x - 1} = (243)^3$

$(3^3)^{2x - 1} = (3^5)^3$

$(3)^{3{2x - 1}} = (3)^{5×3}$

3(2x - 1) = 5 × 3

or 2x - 1 = 5 ∴ x = 3

$(36)^{1/6}=(6^2)^{1/6}$

=$(6)^{2/6}=(6)^{1/3}=√^3{6}$

$2^x = 3^y = 6^{ - z}$

$2^x = 3^y = 6^{ - z}=k$

$2=k^{1/x}; 3=k^{1/y}; 6=k^{-1/z}$

Since, 2×3=6

$k^{1/x}×k^{1/y}=k^{-1/z}$

$k^{1/x+1/y}=k^{-1/z}$

$1/x+1/y=-1/z⇒ 1/x+1/y+1/z=0$