If x0 satisfies the equation log 2(x2-4 x-32)=6 then [ log √3 x0] equals [Note: [ k ] denotes greatest integer less than or equal to k.]

Question

If x0 satisfies the equation log 2(x2-4 x-32)=6 then [ log √3 x0] equals [Note: [ k ] denotes greatest integer less than or equal to k.] (A) 2 (B) 3 (C) 4 (D) 5

Tardigrade Admin · Accepted Answer

x2-4 x-32=64 x2-4 x-96=0 (x-12)(x+8)=0 x=12, x=-8 x ≠-8 ⇒ x2-4 x-32=0 log √3 x0= log √3 12= log 3 144 log 3 81< log 3 144< log 3 243 ⇒ 4< log 3 144<5 ∴ [ log 3 144]=4

Q. If $x_{0}$ satisfies the equation $lo g_{2} (x^{2} - 4 x - 32) = 6$ then $[lo g_{3} x_{0}]$ equals
[Note: $[k]$ denotes greatest integer less than or equal to $k$ .]

2

3

4

5

$x^{2} - 4 x - 32 = 64$
$x^{2} - 4 x - 96 = 0$
$(x - 12) (x + 8) = 0$
$x = 12, x = - 8$
$x \neq = - 8 \Rightarrow x^{2} - 4 x - 32 = 0$
$lo g_{3} x_{0} = lo g_{3} 12 = lo g_{3} 144$
$lo g_{3} 81 < lo g_{3} 144 < lo g_{3} 243 \Rightarrow 4 < lo g_{3} 144 < 5$
$∴ [lo g_{3} 144] = 4$

Q. If x0​ satisfies the equation log2​(x2−4x−32)=6 then [log3​​x0​] equals [Note: [k] denotes greatest integer less than or equal to k.]

2

3

4

5

x2−4x−32=64 x2−4x−96=0 (x−12)(x+8)=0 x=12,x=−8 x=−8⇒x2−4x−32=0 log3​​x0​=log3​​12=log3​144 log3​81<log3​144<log3​243⇒4<log3​144<5 ∴[log3​144]=4

Q. If $x_{0}$ satisfies the equation $lo g_{2} (x^{2} - 4 x - 32) = 6$ then $[lo g_{3} x_{0}]$ equals
[Note: $[k]$ denotes greatest integer less than or equal to $k$ .]

$x^{2} - 4 x - 32 = 64$
$x^{2} - 4 x - 96 = 0$
$(x - 12) (x + 8) = 0$
$x = 12, x = - 8$
$x \neq = - 8 \Rightarrow x^{2} - 4 x - 32 = 0$
$lo g_{3} x_{0} = lo g_{3} 12 = lo g_{3} 144$
$lo g_{3} 81 < lo g_{3} 144 < lo g_{3} 243 \Rightarrow 4 < lo g_{3} 144 < 5$
$∴ [lo g_{3} 144] = 4$