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4)

\(f(x) = \log \left(x + \sqrt{1 +x^2}\right)\\ \begin{array}{rcll} f(-x) &=& \log \left(-x + \sqrt{1 + x^2}\right)\\ &=&\log \left(\dfrac{(-x+\sqrt{1+x^2})(x + \sqrt{1+x^2})}{x+\sqrt{1+x^2}}\right)\\ &=&\log \left(\dfrac{(\sqrt{1+x^2})^2-x^2}{x+\sqrt{1+x^2}}\right)\\ &=&\log \left(\dfrac{1}{x+\sqrt{1+x^2}}\right)\\ &=&\log \left(\left(x + \sqrt{1+x^2}\right)^{-1}\right)\\ &=&-\log\left(x+\sqrt{1+x^2}\right)\\ &=&-f(x)\\ \end{array}\\ \therefore \text{It is an odd function.}\)

3 -  

Solve for x:
((5 sqrt(2) - x)^(1/3) + (x + 5 sqrt(2))^(1/3))^(1/3) = sqrt(2)

Raise both sides to the power of three:
(5 sqrt(2) - x)^(1/3) + (x + 5 sqrt(2))^(1/3) = 2 sqrt(2)

Subtract (x + 5 sqrt(2))^(1/3) from both sides:
(5 sqrt(2) - x)^(1/3) = 2 sqrt(2) - (x + 5 sqrt(2))^(1/3)

Raise both sides to the power of three:
5 sqrt(2) - x = (2 sqrt(2) - (x + 5 sqrt(2))^(1/3))^3

Subtract (2 sqrt(2) - (x + 5 sqrt(2))^(1/3))^3 from both sides:
5 sqrt(2) - x - (2 sqrt(2) - (x + 5 sqrt(2))^(1/3))^3 = 0

5 sqrt(2) - x - (2 sqrt(2) - (x + 5 sqrt(2))^(1/3))^3 = -6 sqrt(2) + 24 (x + 5 sqrt(2))^(1/3) - 6 sqrt(2) (x + 5 sqrt(2))^(2/3):
-6 sqrt(2) + 24 (x + 5 sqrt(2))^(1/3) - 6 sqrt(2) (x + 5 sqrt(2))^(2/3) = 0

Simplify and substitute y = (x + 5 sqrt(2))^(1/3).
-6 sqrt(2) + 24 (x + 5 sqrt(2))^(1/3) - 6 sqrt(2) (x + 5 sqrt(2))^(2/3) = -6 sqrt(2) + 24 (x + 5 sqrt(2))^(1/3) - 6 sqrt(2) ((x + 5 sqrt(2))^(1/3))^2
 = -6 sqrt(2) y^2 + 24 y - 6 sqrt(2):
-6 sqrt(2) y^2 + 24 y - 6 sqrt(2) = 0

Divide both sides by -6 sqrt(2):
y^2 - 2 sqrt(2) y + 1 = 0

Subtract 1 from both sides:
y^2 - 2 sqrt(2) y = -1

Add 2 to both sides:
y^2 - 2 sqrt(2) y + 2 = 1

Write the left hand side as a square:
(y - sqrt(2))^2 = 1

Take the square root of both sides:
y - sqrt(2) = 1 or y - sqrt(2) = -1

Add sqrt(2) to both sides:
y = 1 + sqrt(2) or y - sqrt(2) = -1

Substitute back for y = (x + 5 sqrt(2))^(1/3):
(x + 5 sqrt(2))^(1/3) = 1 + sqrt(2) or y - sqrt(2) = -1

Raise both sides to the power of three:
x + 5 sqrt(2) = (1 + sqrt(2))^3 or y - sqrt(2) = -1

Subtract 5 sqrt(2) from both sides:
x = (sqrt(2) + 1)^3 - 5 sqrt(2) or y - sqrt(2) = -1

(sqrt(2) + 1)^3 - 5 sqrt(2) = 7:
x = 7 or y - sqrt(2) = -1

Add sqrt(2) to both sides:
x = 7 or y = sqrt(2) - 1

Substitute back for y = (x + 5 sqrt(2))^(1/3):
x = 7 or (x + 5 sqrt(2))^(1/3) = sqrt(2) - 1

Raise both sides to the power of three:
x = 7 or x + 5 sqrt(2) = (sqrt(2) - 1)^3

Subtract 5 sqrt(2) from both sides:
x = 7 or x = (sqrt(2) - 1)^3 - 5 sqrt(2)
sqrt(2) - 1)^3 - 5 sqrt(2) = -7:

x = 7              or               x = -7

OMG THX

OMG THX

2.     [  log (a / b) ]^2  +  [ (log ab) ] ^2  =  20         and    (log a - log b) log(ab)  = 8

Note that

log (a / b)  = log a - log b

log (ab)  =   log a + log b

So we have

(log a - log b)^2  + (log a + log b)^2  =  20         and   (log a - log b) ( log a + log b )  = 8

Let log a  = u       Let log b = v

So

(u - v)^2   + ( u + v)^2  = 20                                 and        (u - v) (u + v)  = 8

u^2  - 2uv + v^2   + u^2 + 2uv + v^2  = 20                         u^2  - v^2  = 8     (2)

2u^2 + 2v^2  = 20

u^2 + v^2  = 10    (1)

Add (1)  and (2)    and we have that

2u^2  = 18

u^2  = 9

u = ±3  

So 

u =  log  a  = ± 3

So

l log a  l     =   l ± 3 l  =      3

i *think* this is a math question, well i think its math? but still

Using the change-of base theorem  we have

log (3x + 4)             log (3x + 4)         log (3x + 4)               (1/2)log(3x + 4)

_________    =      __________  =   ______________=    ____________    (1)

log 4                          log 2^2                2 log 2                         log 2

log ( 7x + 8)        log(7x + 8)           log (7x + 8)                (1/3)log(7x + 8)

_________   =  __________  =  _____________ =        _____________   (2)

log 8                    log 2^3                  3 log 2                           log 2

The arithmetic difference must  be   (2) - (1)  =

(1/3) log (7x + 8) - (1/2)log(3x + 4)

__________________________

                  log 2

This implies that

log (x + 2)               (1/3)log(7x + 8) - (1/2)log (3x + 4)              (1/2)log(3x + 4)

________    +        ____________________________   =    ______________

  log   2                                  log 2                                                log 2

;log (x + 2)  +  (1/3)log(7x + 8) - (1/2)log(3x + 4)  =  (1/2)log(3x + 4)

log (x + 2) + (1/3)log(7x + 8)    =   log(3x + 4)

3 log( x + 2)  + log(7x + 8)  = 3log(3x + 4)

log ( x + 2)^3 + log(7x + 8)    =  log (3x + 4)^3

log  [ (x + 2)^3 * (7x + 8)]  = log(3x + 4)^3

Which implies that

(x + 2)^3 (7x + 8)  = (3x + 4)^3

(x^3 + 3x^2*2 + 3x*4 + 2^3) (7x + 8)   =  (3x)^3  + 3* (3x)^2*4 + 3* 3x*4^2 + 4^3

( x^3 + 6x^2 + 12x + 8)(7x + 8)  =  27x^3 + 108x^2 + 144x + 64

7 x^4 + 50 x^3 + 132 x^2 + 152 x + 64  =  27x^3 + 108x^2 + 144x + 64

7x^4 + 23x^3 + 24x^2 + 8x  = 0

x (7x^3 + 23x^2 + 24x + 8)  = 0

x = 0 is one solution

7x^3 + 23x^2 + 24x + 8  = 0

Using the Rational zeroes theorem, I see that x = -1 is also another solution

So....we can use synthetic division to find the residual polynomial

-1  [  7     23     24      8   ]

                -7     -16    -8

       ________________

        7     16       8       0

The   remaining polynomial is     7x^2  + 16x + 8

Using the Quadratic Formula  

x =    -16 ±√ [ 16^2  - 4(7)(8) ]      =          -16  ±√ 32      =      -16 ± 4√2    =     -8 ±2√2

        ____________________                __________          _________        _______

                   2 *7                                            14                          14                       7

The solution  -8 - 2√2

                     _______  ≈  -1.55       which makes two of the original logs undefined  

                           7

So

x  =   -8 + 2 √2

        _________  ≈  -.74

               7

a =  8   b =  2    c  = 2   d  = 7

And their sum  =    19

We can  choose any 2 of 5 postions for the "9s" to occupy.....so

5C2  =   10   five-digit integers

Thank you so much!

We can place all three candies in any box  =  4 ways

We can place  the same candies in any one of the 4  boxes and the different one in any of the remaining 3 boxes

= 4 * 3  = 12 ways

We can place  the different candy and either piece of the same candy in any of the 4 boxes and the other piece of the same candy in any of the remaining 3 boxes = 4 * 3  = 12 ways

[ The same candies are indistinguishable, so choosing either doesn't matter]

We can place the different candy in one of the four boxes and choose any 2 of the remaining 3 boxes to put one piece of each of the same candies in   =  4 * 3C2  =  4 * 3  =  12 ways

So......

4 + 3(12)   = 40 ways

CPhill noooooooo

√3 *∛2  =

(3)^1/2  * (2)^(1/3)  =

(3)^(3/6) * 2^(2/6)  =

(3^3)^(1/6)  *  (2^2)^(1/6)  =

(27)^(1/6) * (4)^(1/6)  =

(27 * 4)^(1/6)    =

(108)^(1/6)   =

⁶√108

\(\sqrt[2]{3}\cdot\sqrt[3]{2}\ =\ 3^{\frac12}\cdot2^{\frac13}\ =\ 3^{\frac36}\cdot2^{\frac26}\ =\ (3^3\cdot2^2)^{\frac16}\ =\ (27\cdot4)^{\frac16}\ =\ (108)^{\frac16}\ =\ \sqrt[6]{108}\)

    sec^2 x

_________________   =

(sec x + 1) (sec x - 1)

   sec^2 x

__________       [note   tan^2 x +  1  = sec^2x     ....so.....sec^2x - 1  = tan^2x  ]

   sec^2x- 1

sec^2 x

_______

tan^2 x

1 / cos^2 x

_____________

sin^2x / cos^2 x

[ 1 /cos^2 x ]  *  [ cos^2x  / sin^2 x  ]

   1

_____

sin ^2x

  1              1

___  *      ____

sinx         sin x

cscx  * cscx  =

csc^2 x

\(\frac{1}{x} + \frac{1}{2y} = \frac{1}{7}\)

[ x + 2y ]              1

_______   =       ___

    2xy                  7

7 [ x + 2y ]  =  2xy

7x + 14y  = 2xy

14y - 2xy  = - 7x

2xy - 14y  =  7x

y [2x - 14 ]  =  7x

y   =                7x

                    _______

                     2x  -14

If

x  = 8   y  =  28         

x = 14  y   = 7

x = 56  y  = 4

We can stop here since if y  = 3   then   

3(2x - 14 = 7x

6x - 42 = 7x

-42  = x

And x will also be negative if y = 2 or y = 1

So.....the minimum value of xy  =  14 *  7  =   98

Ruben has to be 31 years old since she was 30 last year.

31-5 = 26.

Sarah is 26 years old.

~~Hypotenuisance

To solve this problem,

Subtract 10 from both sides.

Then Divide by 2 from both sides.

You will get a = 70.

~~Hypotenuisance

If x = 50,

100 + 2y = 500.

Then, subtract 100 from both sides.

Divide by 2 on both sides.

You will get 200 for y. There is only one solution to y.

~~Hypotenuisance

Wait is this in the correct format?

Thanks, heureka......I like that method  !!!!!