Post mathematic related questions here

[Chaitanya]

May be this will help...
BTW was too lazy type

*tapatalk.imageshack.com/v2/14/11/14/a5bc45d18bf2793a838881c84ec8d5c9.jpg

 

[rst]

Great

Thanks a lot

 

[rst]

Solve for x

2^(x+2)- 6^x -2 [3^(2x +2)]=0

 

[Chaitanya]

so just to be clear the question is

{2^(x+2)} - {6^x} - 2{3^(2x+2)} = 0

 

[rst]

^^ yes question is like that

 

[Niilesh]

Solve for x

2^(x+2)- 6^x -2 [3^(2x +2)]=0

take log both sides then we will get a linear equation in one variable which can be easily solved.

 

[Chaitanya]

^ not that simple.
I couldn't find any satisfactory solution yet..

 

[nomad47]

Here you go


*tapatalk.imageshack.com/v2/15/01/31/a36923181c8cf943b6832d00a29aea23.jpg

 

[Chaitanya]

Buddy on LHS of the equation it will be "log(2^(x+2)-6^x)" & not "log(2^(x+2)) - log(6^x)"

 

[nomad47]

Oops. Damn. My mathematics is getting rusty sorry

 

[rst]

Solve for x

2^(2x+2)- 6^x -2 [3^(2x +2)]=0

4 2^(2x) - 2^x 3^x - 18 3^(2x) =0

Dividing both sides by3^(2x) we get

4 (2/3)^(2x) - (2/3)^x -18 =0

Let (2/3)^x =y

4 y^2 - y -18 =0

Solving
y= 9/4 or -2

(2/3)^x = (3/2)^2

x = -2

 

[rst]

If √(pq) =6 and p,q are positive integers. Then which of the following could not be a value of
(p-q) ?

a) 0
b) -9
c) 8
d) 5

 

[hari1]

Option C:8

 

[rst]

Correct.

In that case we are not getting positive integral value of p,q

 

[rst]

Find the length of the diagonal of the largest possible cube inscribed in a hemisphere of radius 4√2 cm.

A) 12 cm
B) 6 cm
C) 8 cm
D) 4 cm

 

[rst]

Ans is C) 8 cm

 

[rst]

*imageshack.com/i/idJjorfrj

 

[Chaitanya]

Ans is B

 

[Chaitanya]

*tapatalk.imageshack.com/v2/15/02/12/dab353aafa1debeb7511e416c9665413.jpg

 

[Chaitanya]

Well I have a question too..
Its not exactly mathematics but then there is no place else*tapatalk.imageshack.com/v2/15/02/12/2f52b28b05c017fb850a38576d41d2aa.jpg