Writing Assignment 2 Show that loga(x+√(x^2-1)) + loga(x-√(x^2-1))=0 To prove that loga(x+√(x^2-1)) + loga(x-√(x^2-1))=0 one must observe what they have in common. In this case, both logarithms share a base of “loga”. Furthermore, the product rule is to be applied to the sum. In that event, both logarithms will be multiplied. loga(x+√(x^2-1)) (x-√(x^2-1)) One must multiplied “x” times “x” to equal to “x^2”. Next, “x” is to be multiplied by negative “√(x^2-1)”, which equals to negative “x√(x^2-1) ". Then, “√(x^2-1) " is to be multiplied by positive “x” , which equals positive “x√(x^2-1"). Finally, "√(x^2-1)” is to be multiplied by negative “√(x^2-1)”, which equals to “x^2-1” because the square root cancels out. Once the terms have been distribute, …show more content…
For example,"(-x+√(x^2-1))” plus “(x+√(x^2-1))” cancel because they equal to zero. This leaves one with x^2 – (x^2-1), so the minus sign is to be distribute to “(x^2-1)”. This equals to x^2 – x^2+1, and makes x^2 and negative x^2 cancel out because it will also equal to zero. As a result, log(a) is the same as saying log(1), which equals to zero. This is because the “10” to the power of “0” equals to “1”. If f(x)=loga(x), show that (f(x+h)-f(x))/h= loga (1+h/x )^(1/h) , h ≠0 One must plug in “loga” for every “f” found in the equation. (loga(x+h)-loga(x))/h= loga (1+h/x )^(1/h) The quotient rule is to be applied to the difference. This means that on the left side (x+h) will be divide by (x), all over “h”. This also means that “x” is divided by “x”(which equals to 1), and “h” is divided by x, all over h. (loga (1+ h/x ))/h The power rule is to be applied to get rid of “h” from the denominator, which equals to 1/h loga (1+h/x ). According to the power rule , 1/h is to be moved next to (1+h/x ), which equals to loga (1+h/x )^(1/h). In conclusion, (loga(x+h)-loga(x))/h= loga (1+h/x )^(1/h) is true. In addition, “h” cannot equal to zero because it would make the equation
= ½ (a2 + b2) ´ ½ (a2 + b2) eventhough it would be easier to do ab,
To calculate the first derivative, I found the average rate of change of Emmitt Smith’s annual rushing yards from the two years surrounding the year I was deriving. Smith’s yards per year had an increasing slope in the years 1990, ’91, ’94, ’97, ’98, and 2004.
When symbols are wrote this way, it will represent the opposite of the original symbol.
Step Two: The next step, will be to get rid of the square root. So, to get rid of the square root we must square it. Remember with equations, what you do to one side you do it to the other.
Didn’t think so. The order of operations works like this: First anything in the parentheses, then we do the exponents/roots, then any multiplication and division- which is done in that order, then we do
When explaining the topic, I was completely lost and had trouble catching up but as soon as there was a demonstration, I soon caught on and was able to complete each equation with confidence.
6. A small stream flowing at a rate of 8 liters per second has a vertical drop of 1.5 m. What is the maximum power that you can obtain from this stream? (1 liter of water has a mass of 1 kg).
from both sides, leaving us with ½ V2 = GH. When the above equation is
the root to the function, like if it is a parabola with its vertex is placed
Given a point, P(Xn,Yn), on a curve, a line tangent to the curve at P crosses the X axis at a point whose X coordinate is closer to the root than Xn. This X coordinate, we will call Xn+1. Repeating this process using Xn+1 in place of Xn will return a new Xn+1 which will be closer to the root. Eventually, our Xn will equal our Xn+1. When this is the case, we have found the root of the equation.
1.01100 = 2.7048 (1+1/1000)1000 = 1.0011000 = 2.7169 (1+1/10000)10000 = 1.000110000 = 2.7181 (1+1/) = (1+) = 2.7182818284590452353602874713526624977572470936. = e. Works Cited http://betterexplained.com/articles/an-intuitive-guide-to-exponential-functions-e/. http://oakroadsystems.com/math/loglaws.htm http://www.physics.uoguelph.ca/tutorials/LOG/ http://en.wikipedia.org/wiki/Michael_Stifel http://en.wikipedia.org/wiki/John_Napier http://www.ndt-ed.org/EducationResources/Math/Math-e.htm http://www.thocp.net/reference/sciences/mathematics/logarithm_hist.htm http://mathforum.org/dr.math/faq/faq.pi.html
To demonstrate the use of calculus, we will be taking certain examples and solving them
It would be nice to divide both sides by A (to get X=B/A), but remember we can't divide.