2)

Solve for x over the real numbers:
log(x + 1) = log(2) + (log(x))/2

Subtract log(2) + (log(x))/2 from both sides:
-log(2) - (log(x))/2 + log(x + 1) = 0

Bring -log(2) - (log(x))/2 + log(x + 1) together using the common denominator 2:
1/2 (-2 log(2) - log(x) + 2 log(x + 1)) = 0

Multiply both sides by 2:
-2 log(2) - log(x) + 2 log(x + 1) = 0

-2 log(2) - log(x) + 2 log(x + 1) = log(1/4) + log(1/x) + log((x + 1)^2) = log((x + 1)^2/(4 x)):
log((x + 1)^2/(4 x)) = 0

Cancel logarithms by taking exp of both sides:
(x + 1)^2/(4 x) = 1

Multiply both sides by 4 x:
(x + 1)^2 = 4 x

Subtract 4 x from both sides:
(x + 1)^2 - 4 x = 0

Expand out terms of the left hand side:
x^2 - 2 x + 1 = 0

Write the left hand side as a square:
(x - 1)^2 = 0

Take the square root of both sides:
x - 1 = 0

Add 1 to both sides:
Answer: |x = 1

Good proof.

My proof:

\(\text{Let the angle(arccos theta) be }\phi\\ \tan(\arccos\theta) \\ = \tan \phi\\ = \dfrac{\sqrt{1-\theta^2}}{\theta}\\ \text{Let arcsin theta be }\phi_1\\ \cot(\arcsin\theta)\\ =\dfrac{1}{\tan\phi_1}\\ =\tan(90^{\circ}-\phi_1)\\ =\tan \phi\\ =\tan(\arccos\theta)\)

1. Prove that 

\(\tan{(~\arccos{(x)}~)}=\cot{(~\arcsin{(x)}~)}\)

i)

\(\begin{array}{lrcll} & \cos{(\varphi)} &=& x \\ \text{or}& \quad \varphi &=& \arccos{(x)}\\ \end{array}\)

ii)

\(\begin{array}{lrcll} &\sin{(90^\circ-\varphi)} &=& \cos{(\varphi)} = x \\ \text{or}& \quad 90^\circ-\varphi &=& \arcsin{(x)}\\ \end{array}\)

iii)

\(\begin{array}{rcll} \cot{(90^\circ-\varphi)} &=& \tan{(\varphi)} \\ \cot{(\arcsin{(x)})} &=& \tan{(\arccos{(x)})} \\ \end{array} \)

its keyshawn

The variables already have coefficients.......

\(\dfrac{2x-1}{3}=4x\\ 2x - 1 = 12x\\ 10x = -1\\ x = -\dfrac{1}{10}\)

Now we seldom tell jokes here......

Normal users always says 'Math question please' unless it is a math joke question.

Now only a little part of normal users like GingerAle and Nauseated(not seen recently) and the 2 mods: Melody and CPhill may tell math jokes.......

BTW I am kinda new. I registered on April 2016.

EACH 2.5 cm that they are apart is equal to 12 miles....

12 miles is to 2.5 cm as  x miles is to 8 cm

12/2.5 = x/8      12/2.5 * 8 = x = 38.4 miles apart

You know what is SOHCAHTOA?

Sine of an angle is Opposite over Hypotenuse. : SOH

Cosine of an angle is Adjacent over Hypotenuse. : CAH

Tangent of an angle is Opposite over Adjacent. : TOA

Hope this helps :)

Well if you mean a d**k it is here:

https://www.desmos.com/calculator/mnugjqauzc

Well they are true because bananas XD

Equals to 50-49/5x8

=19/40

=0.475

Ans:0.475 or 19/40

Hi DS, I only just saw your question.

It is really good to see you on the forum :)

Are you happy with EP's answers?  

I wish you could visit more often, we still miss you here :)

thanks EP :)

they are false bananas 

Please someone help me i need the answer right now 

133-133=0

Pikachu

5  4

inversine is the same as arc sine.

use asin :)

Take your time...i understand...hope you feel better c: 

Find the distance (departure) to the nearest whole mile between

initial position 50° 00' N 02° 00' W

and 50° 00' N 01° 26' W 

\(\begin{array}{lrcll} \text{Set Latitude } & lat_1 &=& 50^\circ \\ \text{Set Longitude } & lon_1 &=& -2^\circ \\ \text{Set Latitude } & lat_2 &=& 50^\circ \\ \text{Set Longitude } & lon_2 &=& -1^\circ 26' = -\frac{43}{30}^\circ \\ \end{array} \)

\(\begin{array}{lcll} cos(g) = cos(90^\circ - lat1) \cdot cos(90^\circ - lat2) + sin(90^\circ - lat1) \cdot sin(90^\circ - lat2) \cdot cos(lon2 - lon1) \\ \text{with g: Great circle distance} \\\\ \text{Since } cos(90° - a) = sin(a) \text{ and } sin(90° - a) = cos(a) \text{ applies, the formula simplifies to: }\\ cos(g) = sin(lat1) * sin(lat2) + cos(lat1) * cos(lat2) * cos(lon2 - lon1) \\\\ dist = 6378.388 * \pi / 180^\circ \cdot g \\ \text{with dist: distance in km }\\ \end{array} \)

\(\begin{array}{rcll} cos(g) &=& sin(50^\circ) * sin(50^\circ) + cos(50^\circ) * cos(50^\circ) * cos(-1^\circ 26' - (-2^\circ)) \\ cos(g) &=& sin(50^\circ)^2 + cos(50^\circ)^2 * cos(-1^\circ 26' +2^\circ ) \\ cos(g) &=& sin(50^\circ)^2 + cos(50^\circ)^2 * cos(-\frac{43}{30}^\circ +2^\circ ) \\ cos(g) &=& sin(50^\circ)^2 + cos(50^\circ)^2 * cos(\frac{17}{30}^\circ ) \\ cos(g) &=& 0.58682408883 + 0.41317591117 * cos(0.56666666667^\circ ) \\ cos(g) &=& 0.58682408883 + 0.41317591117 * 0.99995109238 \\ cos(g) &=& 0.99997979255 \\ g &=& \arccos{(0.99997979255)} \\ g &=& 0.36424544098^\circ \\\\ dist &=& 6378.388 * \pi / 180^\circ \cdot 0.36424544098^\circ \\ dist &=& 40.5492126920\ km \\ dist &=& \frac{40.5492126920\ km}{ 1.609344\frac{km}{mi}} = 25.1961126347\ mi\\ \end{array}\)

Nah! I am going to change my mind (again) and say "Yes, Melody, you are correct'

I think I even said it in my first answer

DOSE would be the y coordinate      m would be the DOSAGE  'x' would be patient weight and b would be any applicable loading dose

dose = dosage x wt   + load       would make dose the dependent   and  weight the independent variable   

Yah?    

Thanx !     G'Day !

Hope your troubles vanish, HSC.......!!!

We'll miss you.....

Hmmmm.....     the dosage is constant    say  5 mg/kg       the DOSE is variable dependent on weight (which is variable)

so I guess it depends on a slight wording difference whether we are talking   dose   or dosage  (which unfortunately are sometimes used interchangeably) 

For a GIVEN PATIENT  DOSE would be variable between  different medicines based on dosages

For a GIVEN MEDICINE dosage would be constant and patient weight would be variable ....

a conundrum...don't you think?   

Yes Chris wouild be right I read the quesiton incorrectly.

Find the distance (departure) to the nearest whole mile between initial position 50° 00' N 02° 00' W and 50° 00' N 01° 26' W

I read the second one as 26 degrees west INSTEAD OF  1 degree and 26 minutes west.

I will do the last bit a again.  I am assumingthat the earth is a perfect sphere - which it isn't.  

And that the 2 points are both at sea level.

The difference in longitude is 34 seconds.

\(distance = \frac{34}{360*60}*2*\pi*2544.8=25 miles\)

Now wolfram alpha and I agree  

...and here is an online calculator for such things:

40.5 km = 25.17 statute miles  is what it answers.... (21.87 nm)

http://www.movable-type.co.uk/scripts/latlong.html