Aricent Numbers & Decimal Fractions Quiz-1

2^(2x) - 3 * [2^(x) * 2^2] + 32 = 0

2^(2x) - 12 * 2^(x) + 32 = 0

Let's give 2^x a value of y:

y^2 - 12y + 32 = 0

Factor it out:

(y - 8)(y - 4) = 0

y = 4, 8

Now substitute 2^x back for y and solve for each:

2^x = 4
x = 2

2^x = 8
x = 3
----------
Overall:

x = 2, 3

B=ax,c=by,a=cz
b=ax
b=(cz)x
b=(by)xz
b=bxyz
b is cancelled from both side
so, xyz=1

x2+y2= (1+2^1/2)^2+ (1-2^1/2)^2

= (a+b)^2+(a-b)^2

so, we get
1+2+2*2^1/2+1+2-2^1/2
1+2+1+2=6 ans.

put x=2y

> (2y)^y = y^2y

> 2^y * y^y =y^y * y^y

>2^y = y^y

y=2

anything power zero is zero
50^0=1

25^3/2=(5^2)^3/2 then 5^3=125

7^1=unit digit=7

7^2=unit digit=9(7*7=49)last unit digit

7^3=unit digit=3(9*7=63)last unit digit

7^4=unit digit=1(3*7=21)last unit digit

7^5=unit digit=7(1*7=7)last unit digit

it repeats for every 4 cycle

so 71/4=remainder=3 it points 3rd cycle whose unit digit is=3

in same vise 6^anything will be resultant 6;

3^1=3
3^2=9(3*3)
3^3=7(9*3)
3^4=1(7*1)
3^5=3(1*3)
same like 7^71 term thus remainder is 1
multiply 3*6*3=24 .
unit digit is 4

8^-25-8^-26=[(1/8^25)-(1/8^26)]= 7*8^-26

Simple put n=2 in both side so in one side

2^2n-1=2^2*2-1=2^3=8----(1)
.
Another Side 1/8^n-3= 1/8^-1= 1/8^-1=8 ----(2)
.
So compare both equation both are equal than value 2 should be satisfied to equation

2x = 6 -z Þ 2 = 6-z/x -------------(i)
3y = 6 -z Þ 3 = 6-z/y -------------(ii)
Multiplying (i)and(ii)
2x3 = 6-z/x x 6-z/y
6 = 6(1-z/x) *6(1-z/y)
1 = (-z/x) +(-z/y)
1= -z(1/x +1/y)
1/-z = 1/x +1/y
1/x + 1/y + 1/z = 0