describe the difference between the graph of f(x)= log x^2 and f(x)= 2 log x ?
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This is due to a property of log which is that :\nlog(ab)=log(a)+log(b)\nRemark that this is true when well defined (for eg you can't write log(-1*-3)=log(-1)+log(-3) )\n\nSo here, you have log(x^2)=log(x)+log(x)\n\nTherefore, there are no diff. except that log(x^2) exists for all x in R&#92;{0} and 2log(x) is defined for all x strictly positive.
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To graph functions, first get a piece of graph paper or you can free hand the X-Y axis and plot the points between -5 and 5 on both axis.\n\n1) h(x) = 2/x  domain of x (-1, 0, 1, 2)\n\nFind the coordinate pairs for each member of the domain.\nh(-1) = 2/-1  -->  (-1, -2)\nh(0) = 2/0  -->  undefined so no point on the graph\nh(1) = 2/1  -->  (1,2)\nh(2) = 2/2  -->  (2,1)\n\nThe graph will be the three points above.  It is not a line since the domain is limited to the integers listed.\n\n\n2) f(x) = x-1  domain is x/x is a real numbers (which excludes x=0 from the domain, but includes all other real numbers)\n\nSince the equation is in slope-intercept form, i.e.  y = mx + b\nthe slope of the line is m=1 and the intercept on the y-axis is b=-1\n\nYou can draw the line that way, or determine two points on the line like this:\nf(-1) = -1 - 1  -->  (-1,-2)\nf(1) = 1 - 1  -->  (1, 0)\n\nSince the domain is the real numbers except 0, a solid line is drawn to extend beyond both points.  Either way you draw the line, make an open point (ie. a circle) on the line where x=0.  Don't draw the line through the open point.  The line is terminated with arrows to indicate that it extend to infinity in both directions
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log(base x) y=[log(base 10)y]/[log(base10)x]. I hope you can take up from here!
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[ 1/4 LOG (16) ] - [ 2 LOG (6) ] + [ LOG (3) ]\n\n[ LOG ((16)^(1/4)) ] - [ LOG ((6)^2) ] + [ LOG (3) ]\n\n[ LOG (2) ] - [ LOG (36) ] + [ LOG (3) ]\n\n...........................the first two LOGS have the same base of 10, \n...........................so when subtracting LOGS we divide.\n\n[ LOG (2/36) ] + [ LOG (3) ]\n\n[ LOG ((2divided by 2) / (36 divided by 2)) ] + [ LOG (3) ]\n\n[ LOG (1/18) ] + [ LOG (3) ] \n\n...........................the above two LOGS have the same base of 10, \n...........................so when adding LOGS we multiply.\n\nLOG ((1/18) times (3))\n\nLOG ((1/18) times (3/1))\n\nLOG (3/18)\n\n..................................then reduce the fraction by dividing by 3/3.\n\nLOG ((3 divided by 3) / (18 divided by 3))\n\nLOG (1/6)   \n\nLOG ((6)^(-1))\n\n..........an exponent can be moved and multiplied to the front of the LOG.\n\n(-1) times [ LOG (6) ]\n\n - LOG (6)\n\n\nTherefore, \n\n[ 1/4 LOG (16) ] - [ 2 LOG (6) ] + [ LOG (3) ] =     - LOG (6)
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log_2(X) + log_2(X+5) = 1\nlog_2(X * (X+5)) = 1\n2 ^ log_2(X * (X+5)) = 2^1\nX * (X+5) = 2\nX^2 + 5*X = 2\nX^2 + 5*X - 2 = 0\nX = (-5 +- sqrt(5^2 + 4*1*2)) / 2\nX = (-5 +- sqrt(33)) / 2\n\nSo, two possible answers are:\nX1 = (-5 + sqrt(33)) / 2\nX2 = (-5 - sqrt(33)) / 2\n\nAlso, logarithms have sense only if these inequality are true:\nX > 0\nX + 5 > 0\n\nX2 < 0, then X1 is the only possible answer.\n\nFinally:\nX = (sqrt(33) - 5) / 2
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f(x,y) = x^2*y*(e^(-x^2-y^2))\n\nfx(x,y) = 2x*y*e^(-x^2 - y^2)(1-2x^2) this is wrong:\nf_x(x,y) = x^2y ( e^(-x^2-y^2)) (-2x) + 2xye^(-x^2-y^2))\n= 2xye^(-x^2-y^2) [-x^2 + 1] =0\nif x=0 or y=0 or x^2 =1, \nx=0, y=0, x=1 or x=-1\n\nfy(x,y) = x^2*e^(-x^2 - y^2)(1-2y^2) \nf_y(x,y)=  x^2 y (e^(-x^2-y^2)) (-2y) +  x^2(e^(-x^2-y^2))\n=x^2e^(-x^2-y^2) [-2y^2 + 1] \nx=0, y=1/sqrt(2),y=-1/sqrt(2)\n\nso the critical points are: (0,y)\n(0,0), \n(1,1/sqrt(2)), (1,-1/sqrt(2))\n(-1,1/sqrt(2)),(-1,-1/sqrt(2))\ns
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you can change the base...\nln x = (log2 (x))(ln 2)
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F'(x) is the derivitive of F(x), which means it describes the change in F with respect to x.  Another way to say this is that F'(x) is the slope of F(x).  \n\nTo sketch the graph of F'(x), look at your graph of F(x).  Pick a point on that graph (maybe the first point). See what the slope (remember rise over run?) of F(x) is at that point.  That slope is the derivitive F' for that value of x.  Plot this on your F'(x) skecth.  Go the the next point and repeat.
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Log(a^b)= b*Log(a)\nLog(ab) = Log(a)*Log(b)\n\na) Log147= ?\n147= 3*7*7\n=> Log147= Log3 * (Log7)^2\n=> Log147= 2yw\n\nb) Log28= ?\n28= 2*2*7\n=> Log28= (Log2)^2 * Log7\n=> Log28= 2xw
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Here is a tutorial...you can change any of the parameters that you mentioned and look at the graph as it is made and changed...really cool if you ask me.\nhttp://www.analyzemath.com/logfunction/logfunction.html
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Nope,\n\n   The log function blows up to infinity.  (If you don't believe this, give me a value for y=f(x) = K * log(x).  Ha ha! I can give you a value for x such that f(x) is greater than the value you gave me! )\n\n  Your pal from Calc I\n   cuntsucker
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A Natural log. It is expressed as: ln()\nFor example: ln(e^2) = 2
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(1)\nUse log graph paper and plot a couple of known values onto the graph paper.  Then read off the required values and work out your answer.\n = log 15 / log 17\n = 1∙176 912 59 / 1∙230 448 921\n = 0∙955 822 902\n\n(2)\nUse a log table book.
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Do you mean \n\nlog (base n) (x) = 1/(log (base x) (n))\n\nWell let log (base n) (x) = a\nTherefore x = n^a\n\nTake logs (base x)\nSo 1 = log (base x) (n^a)\n= a * log (base x) (n)\nBut a = log (base n) (x)\nTherefore log (base n) (x) * log (base x) (n) = 1\nie log (base n) (x) = 1/(log (base x) (n)) ............. QED
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d (ln(x)) / dx = 1/x\n\nThe derivative of the natural log of x with respect to x equals 1 / x
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Remember that with logarithms, exponents within the log become the coefficient outside the log.\n\n1.) logc (sqrt x)\n= logc (x^1/2)\n= (1/2) logc x\n= (1/2) * 1.2\n= 0.6\n\nIf numbers are being multiplied within the log, they can be split up to the sum of the log of the numbers.  It's a similar rule to that in problem 1, as you can imagine:\n\nlog (x^2) = log (x * x) = log x + log x = 2 log x\n\n2.) logc xy\n= logc x +logc y\n= 1.2 + 3.9\n= 5.1
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First let me assert that \n  log(base b)(a)=1/log(base a)(b) as follows:\n\n let log(base a)b=z\n   a^z=b\n   a=b^(1/z)  . Now sub back for z giving\n   a=b^(1/log(base a)(b))  and convert to log form of base b\n 1)   log(base b)(a)=1/log(base a)(b)  \n\n Now follow this: \n    let  log(base c)(x)=y\n        c^y=x   Now take log(base a) of both sides\n    ylog(base a)c=log(base a)x  and  sub back for y\n log(base a)x  = log(base c)x{log(base a)c} and\n use 1) to convert log(base a)c giving\n   2)  log(base a)x  = log(base c)x/{log(base c)a.\n\n     Lastly, into 2)  substitute X^3+1 for x,  4 for a and 10 for c\n\nand we have \n  log(base 4)(x^3+1)  = log(base 10)(x^3+1)/{log(base 10)4}