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Here are some examples:

To find  log₁₀(100)  ,   type in:     log(100)

To find  log_e(100)   ,   type in:     ln(100)

To find  log₃(100)   ,   type in:    log(100,3)

Nope, but I can try. 

5 )

\($f(x)=x+1$\)

\($\overbrace{f(f(\cdots(f}^\text{2015 total \textit{f}s}(x))\cdots)=0$?\)

Note  that  if  we had    x +  1.......then...

f ( f ( -2) )   =  f ( -1)  =  0

And

f ( f ( f (-3) ) )  =  f (f (-2)  )  =  f (-1)   =  0

So....this implies that  x  =  -2015

(1/2)   l  3x  -  5   l    =  x  + 1

Multiply both sides by 2

l 3x  -  5 l  =   2x  + 2

We have two equations here

3x  -  5     =    2x  +  2                  and           3x  -  5  =   -[ 2x + 2]

subtract 2x from both sides                            simplify

add 5 to both sides                                       3x  - 5    =  -2x  -  2

x  = 7                                                           add 2x , 5  to both sides

                                                                             5x  =   3

                                                                        divide both sides by 5

                                                                               x  = 3/5

So......the solutions  are  x  = 7  and x  = 3/5

140, a,  45/28

Let r be the common ratio...so

140r  = a      ⇒  r  =  a /140

So..this implies that

a (a / 140)   =  45/28

a^2  =  (140*45) / 28

a^2   = 5 * 45

a^2  =  225   

a =  15

CANADA!!!

Sqrt[ (45/28) / 140] =3/28 - Common ratio.

3/28 x 140 = 15 - the 2nd term "a"

\( $\sqrt {14 + 8\sqrt {3}}$\)

In the form   √   [ a  + b√c ]      we  can  find a simplification  if

(b/2)^2  + ( √c ) ^2    =  a        

And  the simpification becomes  √ [   ( (b/2) + √c )^2  ]

(4)^2  + 3  =   19      isn't  true.....but notice that  we can  express this as

√ [ 2  ( 7  + 4√3)  ]   =  √2 * √ [ 7 + 4√3 ]

Then  considering the second radical....

(4/2)^2  + (√3)^2   =  2^2 + 3   =  7  is true

So  we can write

√2  * √ [   ( (4/2)  + √3 ) ^2 ]  =

√ 2  *  √ [ ( 2 + √3)^2 ]  =

√2 [ 2 + √3]  =

2√2  + √6  =

√8  +  √6

day 2    3 x 2 cells

day 4    3 x 2 x 2 cells

day 6    3 x 2 x 2 x 2 cells

day 8    3 x 2 x 2 x 2 x 2 cells = 48 cells

3*2=6*2=12*2=24*2=48

There are 48 cells after 8 days.

6666/60 = 111 minutes   6 seconds

111 minutes = 1 hr 51 minutes

3:00:00 + 1:51:06   =   4:51:06 PM

A coupe of physics equations:

1/f = 1/do + 1/di      do = object distance   di = image distance

and    M = -di/do       or  do = -di/M    (substitute into the first equation)

1/20 = 1/(-di/-4) + 1/di

1/20 = 4/di + 1/di

1/20 = 5/di      di = 5(20) = 100 cm

M= -di/do

-4 = -100/do     do = 25 cm

Distance between do and di =    100 + 25 = 125 cm

3)

Let   x  =  √[ 6 + √ [6 + √6 +.... ] ]     square both sides

x^2  =   6 +    √[ 6 + √ [6 + √6 +.... ] ] 

x^2  =  6  + x       rearrange as

x^2  - x  -  6  = 0       factor

(x - 3)  ( x + 2)  =  0

Setting  each factor to  0  and solving for x  gives that   x  = 3   or x  = -2

Reject the second

So.....x  =   3  =  the   value

1) 

 \( \[ f(x) = \dfrac {1}{1 - \dfrac {1}{1 - \dfrac {1}{1 - x}}}, \] \)

\($(f(f( - 2)))^{ - 2}$\)

f ( -2)  =  -2

And   f ( f(-2)) =  f (-2)   =  -2

So

( f ( f(-2) ))^-2  =  (-2)^-2  =    1  / (-2)²  =    1/4

I was looking at this question the other day and thinking of an easy way to solve it....then I saw CPHIL's answer which was different than my thinking, but I wil use his graph diagram rather than uploading my hand drawings.....

The short thick black line is = 3  (given in the question)   the slope of the blue line is - 12/5  (given)....  Now let's rotate the triangle so the blue line is verticle ..like this:

Now the thin black line (hypotenuse, h) that we are looking for has a slope of 12/5 

We can find the length of the blue line, b, by  TAN = opp/adj = 12/5 = b/3     so b = 36/5   

Now, use Pythag theorom to find the hypotenuse (the distance on the y axis betweeen the two parallel lines)

h^2= (36/5)^2  + 3^2    results in h= 7.8

we need TWO h's to find the distance between the TWO lines  parallel to the given line:     2 x 7.8 = 15.6 

Hope this isn't too confusing !

Let $d$ and $q$ be the roots of the quadratic equation $x^2+7x-4=0$. Find the value of $2d^2+19d+5q-8$.

x^2  +  7x  -  4  =  0

The roots are

[-7 + √ [ 7^2  + 16 ] ] / 2  = [-7 + √ [ 65 ] ] / 2    =    d

And

[-7 - √ [ 7^2  + 16 ] ] / 2  =   [  - 7 - √65] / 2    =  q

So......

2d^2  +  19d +  5q  - 8      = 

(1/2) [ 49 -  14√ 65  + 65]  +  (19 / 2) [-7 + √ [ 65 ] ]  + (5/2)  [  - 7 - √65]  - 8  =

49/2  -  7√ 65  +  65/2   - 133/2 +  (19/2)√ 65  -  35/2  - (5/2)√ 65  - 8   =  

[ 49  + 65  -  133 - 35 -16] / 2   

[ 114   -  184  ] / 2

[-70] / 2  =

-35

BTW......the results are the same no matter which root we assign to d, q

how about (10+10)/(10-10) = 20/0

OH SHOOT I MIGHT END THE WOR-

Well I wasn't really looking for a solution, I just wanted to know what LN was...thanks anyway!

cant use log. much more basic.

log 10 + log 10 + log 10 + log 10 = 4, and 10 - log 10 - log 10 - log 10 = 7 perhaps?

:P

log is a function, not a number.

Still 0.

In this case, you think of a number that when it multiplies itself, it gives out 0.

0 * 0 = 0, so \(\sqrt0\) = 0.

Determine the number of positive integers a less than 12 such that the congruence 

\(ax\equiv 1\pmod{12}\)

has a solution in x.

\(\begin{array}{ll} & ax \equiv 1\pmod{12} \\ \text{or} \\ & ax-1 = y\cdot 12 \\ \text{or} \\ & \mathbf{ax -12y = 1} \\ \end{array}\)

\(\begin{array}{|lll|} \hline \text{The linear Diophantine equation takes the form ax + by = c, where a, b and c are given integers.}\\ \text{The solutions are described by the following theorem:}\\ \text{This Diophantine equation has a solution (where x and y are integers)} \\ \text{if and only if c is a multiple of the greatest common divisor of a and b.} \\ \hline \end{array}\)

Let c = 1

\(ax + by = 1 \)

Let b = -12

\(ax-12y = 1\)

There is a solution, if c=1 is a multiple of the gcd(a,-12).   gcd = greatest common divisor

Or gcd(a,-12) | 1,

\(\Rightarrow \) gcd(a,-12) = 1, because 1 is the only divider of 1!

\(\begin{array}{|r|c|} \hline a & gcd ( a,-12)=1 \ ? \\ \hline {\color{red}1} & {\color{red}1}\\ 2 & 2\\ 3 & 3\\ 4 & 4\\ {\color{red}5} & {\color{red}1}\\ 6 & 6\\ {\color{red}7} & {\color{red}1}\\ 8 & 4\\ 9 & 3\\ 10 & 2 \\ {\color{red}11} & {\color{red}1} \\ \hline \end{array}\)

Number of positive integers a less than 12 such that the congruence
\(ax\equiv 1\pmod{12}\)

are 1, 5, 7, 11

Ohh I see, they're asking for a value. 

Thanks everyone! 

Simplify the expression \[\frac{6}{\sqrt{36} + \sqrt{27}} + \frac{6}{\sqrt{27} + \sqrt{18}} + \frac{6}{\sqrt{18} + \sqrt{9}}.\]

\(\begin{array}{|rcll|} \hline && \frac{6}{\sqrt{36} + \sqrt{27}} + \frac{6}{\sqrt{27} + \sqrt{18}} + \frac{6}{\sqrt{18} + \sqrt{9}} \\\\ &=& 6\cdot \left( \frac{1}{\sqrt{36} + \sqrt{27}} + \frac{1}{\sqrt{27} + \sqrt{18}} + \frac{1}{\sqrt{18} + \sqrt{9}} \right)\\\\ &=& 6\cdot \left( \frac{\sqrt{36}-\sqrt{27}}{(\sqrt{36}+\sqrt{27})(\sqrt{36}-\sqrt{27})} + \frac{\sqrt{27}-\sqrt{18}}{(\sqrt{27}+\sqrt{18})(\sqrt{27}-\sqrt{18})} + \frac{\sqrt{18}-\sqrt{9}}{(\sqrt{18}+\sqrt{9})(\sqrt{18}-\sqrt{9})} \right)\\\\ &=& 6\cdot \left( \frac{\sqrt{36}-\sqrt{27}}{36-27} + \frac{\sqrt{27}-\sqrt{18}}{27-18} + \frac{\sqrt{18}-\sqrt{9}}{18-9} \right)\\\\ &=& 6\cdot \left( \frac{\sqrt{36}-\sqrt{27}}{9} + \frac{\sqrt{27}-\sqrt{18}}{9} + \frac{\sqrt{18}-\sqrt{9}}{9} \right)\\\\ &=& 6\cdot \left( \frac{\sqrt{36}}{9} -\underbrace{\frac{\sqrt{27}}{9} + \frac{\sqrt{27}}{9} }_{=0} -\underbrace{\frac{\sqrt{18}}{9} + \frac{\sqrt{18}}{9} }_{=0} -\frac{\sqrt{9}}{9} \right)\\\\ &=& 6\cdot \left( \frac{\sqrt{36}}{9} -\frac{\sqrt{9}}{9} \right)\\\\ &=& \frac{6}{9}\cdot \left( \sqrt{36} - \sqrt{9} \right)\\\\ &=& \frac{6}{9}\cdot \left( 6-3 \right)\\\\ &=& \frac{6}{9}\cdot 3 \\\\ &=& \frac{6}{3} \\\\ &\mathbf{=}& \mathbf{2} \\ \hline \end{array}\)