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I will call the distance from B to C x.

Since all of the points are equal distance from eachother, $FG = 3x \operatorname{and} HI = 4x$.

Area of a trapezoid = $^{1}\!\!/\!_{2}(b_1+b_2)\times h$, where h = height, b₁ = length of bottom, and b₂ = length of top.

Area of FGIH = $7x(\tfrac{\sqrt{3}}{2}x)=\tfrac{7\sqrt{3}}{2}x^{2}$

Area of ABC = $x(\tfrac{\sqrt{3}}{2}x)=\tfrac{\sqrt{3}}{2}x^{2}$

Area ratio of FGIH to ABC = $\tfrac{7\sqrt{3}}{2}x^{2}:\tfrac{\sqrt{3}}{2}x^{2}$ --> $7:1$

Your post just reminded me of something - 

When writing a number in base 3, all numbers that are multiples of 2, when taking the sum of their digits in base 3, were even.

I tried this out with 4 and 3, and it worked there as well.

I then figured out a pattern that I forgot to ask about - 

When writing a number in base x, do all numbers that are multiples of $x-1$, when taking the sum of their digits, make another multiple of $x-1$?

This all started because of the trick that I used to figure out if a number was a multiple of 9 - sum up the digits until they are one. If it is 9, then it is a multiple of 9.

Deja vu...

I once did that, then realized what happened when I gave a thumbs down. I hope I got everyone I put down and gave them a thumbs up instead! 

See this USGS calculator to find the difference between "Intensity" and "Energy Release" :

https://earthquake.usgs.gov/learn/topics/calculator.php

x - 3y   =   9

                           Subtract  x  from both sides.

-3y   =   -x + 9

                           Divide both sides by  -3 .

y   =   x/3 - 3       The slope of this line is  1/3 .

y   =   -3x + 3       The slope of this line is  -3 .

Since the slopes of the two lines are negative reciprocals of each other, the lines are perpendicular.

Thanks.....it's yours

Hey! I like that way! 

Solve for x:
x^(log(x)/(log(sqrt(x)))) = 9

Simplify and substitute y = sqrt(x).
x^(log(x)/(log(sqrt(x)))) = sqrt(x)^4
 = y^4:
y^4 = 9

Taking 4^th roots gives sqrt(3) times the 4^th roots of unity:
y = -sqrt(3) or y = -i sqrt(3) or y = i sqrt(3) or y = sqrt(3)

Substitute back for y = sqrt(x):
sqrt(x) = -sqrt(3) or sqrt(x) = -i sqrt(3) or sqrt(x) = i sqrt(3) or sqrt(x) = sqrt(3)

Raise both sides to the power of two:
x = 3 or sqrt(x) = -i sqrt(3) or sqrt(x) = i sqrt(3) or sqrt(x) = sqrt(3)

Raise both sides to the power of two:
x = 3 or x = -3 or sqrt(x) = i sqrt(3) or sqrt(x) = sqrt(3)

Raise both sides to the power of two:
x = 3 or x = -3 or x = -3 or sqrt(x) = sqrt(3)
Raise both sides to the power of two:
x = 3    or x = -3     or x = -3      or x = 3

Domain: {x element R : 2 3 + sqrt(5)} (assuming a function from reals to reals)

y   =   2x² + kx + 6

                                        We are given that   y = -x + 4  , so we can substitute  -x + 4  in for  y .

-x + 4  =  2x² + kx + 6

                                        Add  x  to both sides of the equation.

4   =   2x² + kx + x + 6

4   =   2x² + (k + 1)x + 6

                                        Subtract  4  from both sides of the equation.

0   =   2x² + (k + 1)x + 2

This equation will have one solution when...

(k + 1)² - 4(2)(2)   =   0

(k + 1)² - 16   =   0

(k + 1)²   =   16

k + 1   =   ± 4

k   =   ± 4 - 1

k   =   4 - 1   =   3          or          k   =   -4 - 1   =   -5

Alternatively 

6x - y + k = 0 then substitute (4,10)

ie k = -14

therefore equation becomes 6x - y -14 = 0 ie y = 6x - 14

2x^2+kx+6 = -x+4

2x^2+ x(k+1) +2 = 0

(2x+     )(x+   ) = 0     the only two factors which multiply to '2' are 1 and 2   or -1 and -2

(2x+1) (x+2)              results in TWO values of x (-1/2 and -1)that would satisfy the system of equations...no good.

(2x+2)(x+1) = 0         would result in ONE values for 'x' :   (-1)    ....we want only ONE (Per the question)

                                     so this would make k+1 = 4   or k=3

Multiply this out to get     2 x^2 + 4x +2 =0        so  k+1 = 4   .... or  k=3

(2x-2)(x-1) =0                works  for only   x =1

    multiply it out     2x^2 -4x+2       means       k+1 = -4       or k=-5

(2x-1)(x-2)      has two solutions    x = 1/2   and x= 2     no good

SO k = 3  or -5    results in only one solution for the system of equations.

Graphically:

The functions are inverses of each other.

The graphs of functions are symmetric to each other over the line  y = x .

First we need to find the slope of the line...

y - 5  =  6x - 10

                            Add  5  to both sides of the equation.

y  =  6x - 5           Now that this is in slope-intercept form, we can see that its slope is  6 .

The slope of a line that is parallel to this will also be  6 .

Using the point  (4, 10)  and a slope of  6  , the equation of our line in point-slope form is...

y - 10   =   6(x - 4)

                               Distribute the  6  to both terms in parenthesees.

y - 10   =   6x - 24

                               Add  10  to both sides of the equation.

y   =   6x - 14          This is the equation in slope-intercept form.

Solve for x over the real numbers:
log(3, log_0.5^2(x) - 3 log(0.5, x) + 5) = 2

log(3, log_0.5^2(x) - 3 log(0.5, x) + 5) = log(5 + 4.32809 log(x) + 2.08137 log^2(x))/log(3):
log(5 + 4.32809 log(x) + 2.08137 log^2(x))/log(3) = 2

Multiply both sides by log(3):
log(5 + 4.32809 log(x) + 2.08137 log^2(x)) = 2 log(3)

2 log(3) = log(3^2) = log(9):
log(5 + 4.32809 log(x) + 2.08137 log^2(x)) = log(9)

Cancel logarithms by taking exp of both sides:
5 + 4.32809 log(x) + 2.08137 log^2(x) = 9

Divide both sides by 2.08137:
2.40227 + 2.07944 log(x) + log^2(x) = 4.32408

Subtract 2.40227 from both sides:
2.07944 log(x) + log^2(x) = 1.92181

Add 1.08102 to both sides:
1.08102 + 2.07944 log(x) + log^2(x) = 3.00283

Write the left hand side as a square:
(log(x) + 1.03972)^2 = 3.00283
Take the square root of both sides:
log(x) + 1.03972 = 1.73287 or log(x) + 1.03972 = -1.73287

Subtract 1.03972 from both sides:
log(x) = 0.693147 or log(x) + 1.03972 = -1.73287

Cancel logarithms by taking exp of both sides:
x = 2. or log(x) + 1.03972 = -1.73287

Subtract 1.03972 from both sides:
x = 2. or log(x) = -2.77259

Cancel logarithms by taking exp of both sides:
x = 2                          or                     x = 0.0625

Domain:{x element R : x>0} (all positive real numbers) (assuming a function from reals to reals)

lynx7, please don't copy an answer without giving credit.

If you find the answer on the forum already, just put a link to it...like this...

Solve for x over the real numbers:

(2 log(x^3))/log(4 x) = (5 log(x))/log(2 x)

Cross multiply:

2 log(2 x) log(x^3) = 5 log(x) log(4 x)

Subtract 5 log(x) log(4 x) from both sides:

2 log(2 x) log(x^3) - 5 log(x) log(4 x) = 0

Transform 2 log(2 x) log(x^3) - 5 log(x) log(4 x) into a polynomial with respect to log(x):

log^2(x) - 4 log(2) log(x) = 0

Factor log(x) and constant terms from the left hand side:

-log(x) (4 log(2) - log(x)) = 0

Multiply both sides by -1:

log(x) (4 log(2) - log(x)) = 0

Split into two equations:

4 log(2) - log(x) = 0 or log(x) = 0

Subtract 4 log(2) from both sides:

-log(x) = -4 log(2) or log(x) = 0

Multiply both sides by -1:

log(x) = 4 log(2) or log(x) = 0

4 log(2) = log(2^4) = log(16):

log(x) = log(16) or log(x) = 0

Cancel logarithms by taking exp of both sides:

x = 16 or log(x) = 0

Cancel logarithms by taking exp of both sides:

x = 16               or                  x = 1

Domain:  {x element R : 0 1} (assuming a function from reals to reals)

Solve for x:

(log(27) x^2)/log(9) = x + 4

Subtract x + 4 from both sides:

(log(27) x^2)/log(9) - x - 4 = 0

Divide both sides by log(27)/log(9):

-(log(9) x)/log(27) + x^2 - (4 log(9))/log(27) = 0

Add (4 log(9))/log(27) to both sides:

x^2 - (log(9) x)/log(27) = (4 log(9))/log(27)

Add (log^2(9))/(4 log^2(27)) to both sides:

-(log(9) x)/log(27) + x^2 + (log^2(9))/(4 log^2(27)) = (log^2(9))/(4 log^2(27)) + (4 log(9))/log(27)

Write the left hand side as a square:

(x - log(9)/(2 log(27)))^2 = (log^2(9))/(4 log^2(27)) + (4 log(9))/log(27)

Take the square root of both sides:

x - log(9)/(2 log(27)) = sqrt((log^2(9))/(4 log^2(27)) + (4 log(9))/log(27)) or x - log(9)/(2 log(27)) = -sqrt((log^2(9))/(4 log^2(27)) + (4 log(9))/log(27))

Add log(9)/(2 log(27)) to both sides:

x = sqrt((log^2(9))/(4 log^2(27)) + (4 log(9))/log(27)) + log(9)/(2 log(27)) or x - log(9)/(2 log(27)) = -sqrt((log^2(9))/(4 log^2(27)) + (4 log(9))/log(27))

Add log(9)/(2 log(27)) to both sides:

| x = sqrt((log^2(9))/(4 log^2(27)) + (4 log(9))/log(27)) + log(9)/(2 log(27)) or x = log(9)/(2 log(27)) - sqrt((log^2(9))/(4 log^2(27)) + (4 log(9))/log(27))

x = -4/3                 and                   x = 2

Domain: {x element R : 0 1} (assuming a function from reals to reals)

Since the question only mentions magnitude and intensity, I guessed that a "stronger" earthquake meant that it has a higher intensity.

Let  I₁  be the intensity of the earthquake with magnitude  8 .

Using the formula given in your question...

8   =   log( I₁/S )          Solve for  I₁ .

                                   Take the inverse of the log of both sides.

10⁸   =   l₁/S

                                   Multiply both sides by  S .

10⁸S   =   l₁

Let  I₂  be the intensity of the earthquake with magnitude  6 .

6   =   log( l₂/S )

10⁶S   =   l₂

The question is...    l₁   equals what times  l₂  ?

l₁  =  x * l₂

10⁸S   =   x * 10⁶S

[ 10⁸S ] / [ 10⁶S ]   =   x

10⁸ / 10⁶   =   x

100   =   x

So...

l₁  =  100  *   l₂

An earthquake with a magnitude 8 is 100 times more intense than an earthquake with magnitude 6 .

Sorry, but that is wrong!

ln(121) =4.7958

5*8^(4x) =376 divide both sides by 5

8^(4x) = 75.2 take the log of both sides

4x*log(8) = Log(75.2) divide both sides by 4*log(8)

x =Log(75.2) /4*log(8)

A =(0.97)^t -Exponential decay

Y =3.7(0.2)^t -Exponential decay

Y =9/11(4)^t -Exponential growth

P =7/9(5/4)^t Exponential growth

V =0.8(9)^t - Exponential growth

P =0.9(8.3)^t - Exponential growth

g(x) =2.1(x) - This is neither - it is linear growth.

I don't quite understand the formula given in the question for "standard earthquake" !!

This is the most modern updated formula used by the USGS(United States Geological Survey):

Energy release in Megatons =10^((1.5 * M)+4.8) / (4.184E15), where M = Magnitude.

So, 6 Magnitude earthquake will release: 10^((1.5*6) +4.8) /4.184E15 =15 Kilotons - or the size of the atomic bomb dropped on Hiroshima in 1945.

But, 8 Magnitude earthquake will release: 10^((1.5*8 +4.8) /4.184E15 =15 Megatons !!!

15 Megatons / 15 Kilotons =15,000,000 / 15,000 =1,000x more poweful is an 8 Magnitude over 6 Magnitude earthquake. The old "Richter Scale" was simply a power of 10. So, using that old scale, the difference between 8 Magnitude and 6 Magnitude earthquake would have been: 10^(8 - 6) =10^2, or 100 times more poweful. But, that is no longer used to measure the Magnitudes of earhquakes. You may go to the USGS website and learn about the moderm formulas used to measure earthquakes.

Note: You should refer this to your teacher.

From the graph   f(x)   (which is 'y') is a constant value from   x =   -6  to 0

5 = Log_0.9(0.59049)

0.9^5 = 0.59049