For the exponential function ex and logarithmic function log x, graphically show the effect if x is doubled.?
827182
for a graph, go to http://www.calculator.com/calcs/GCalc.html\n\ntype in \n\ne^x\nclick enter\ne^(2x)\nclick enter\n\nand then seperately\n\nlog(x)\nclick enter\nlog(2x)\nclick enter\n\nby the way\n\ne^(2x) = (e^x)^2\n\nso in comparison to e^x, the problem was squared\n\nas for\n\nlog(2x) = log(2) + log(x)\n\nso in comparison to log(x)\n\nlog(2x) was moved to the the left log(2) units
697108
e^(1+lnx) = e^1 * e^lnx = e*x = ex\nder(ex) = e+c
813587
2x=x(1.145)^t\nlog2+logx=logx+t log1.145\nt log 1.145t=log2\nt=log2/log1.145\n=0.3010/0.0588\n=5.2 years approx
669001
e^f(x) = x is the best answer i can give you.
956834
For log(x+8) to be log x +log 8 ,x+8 must be equal to 8x\nWe can write it algebrically as log(x+8)=logx+log8=log(8x)\n=>8x=x+8\n=>7x=8    or,x=8/7
994994
If you are saying \n\n1.) (log(.78))/(log(2)) = -.35845397... or log(2)(.78)\n2.) (log(.80))/(log(2)) = -.32192809... or log(2)(.80)\n\nbut if you mean\n\n1.) (log(78))/(log(2)) = 6.2854022... or log(2)(78)\n2.) (log(80))/(log(2)) = 6.32192809... or log(2)(80)
477145
log(2x+1) - log5  = log x\nlog(2x+1) - log x = log 5\nlog((2x+1)/x)  = log 5\n(2x+1)/x  = 5\n2x+1 = 5x\n3x = 1\nx= 1/3
780572
First let's remember some rules of logs.\nThe log of product is the sum of the logs.  That is,\nlob(a*b) = log(a) + log(b)\nThe log of a quotitent is the difference of them.  That is, \nlog(a/b) = log(a) - log(b)\nAnd finally, if you have an exponent, you can bring it in front of the log.\nlog(a^n) = n*log(a)\nRemember that you can only take the log of positive numbers.\nln[ x^2 / [sqrt(x) (2+x)^2] ]\n= ln(x^2) - ln[ sqrt(x) (2+x)^2]\n= 2ln(x) - [ln(sqrt(x)) + ln(2+x)^2]\n= 2ln(x) - (1/2)ln(x) - 2ln(2+x)\nYou can combine 2ln(x) - (1/2)ln(x)\n2ln(x) - (1/2)ln(x) = (2 - 1/2)ln(x) = (3/2)ln(x)\nAnswer: (3/2)ln(x) - 2ln(2+x)\nNote: sqrt(x) = x^(1/2)
338108
A Natural log. It is expressed as: ln()\nFor example: ln(e^2) = 2
84854
the exponential (both b>1 and 0<b<1) doesn't intercept the x axis: in fact the x axis is a horizontal asymptote for the function (asymptote = which doesn't touch, by ethimology).\n\nAnyway, if you want an algebrical answer, substitute -1 to x in the function: you get y=b^-1 --> y=1/b\nThe reciprocal of any number is never 0.
861527
Stop, stop, stop. If it was f(x)=2logx, f(x) cannot be = to (1/log2)*logx, as  1/log2 is not = 2.\nDon’t hurry, put your question correctly with necessary details and don’t save on ()-s in formulas.
808044
Take the log of both sides:\nlog(0.01) = log(2.2^-t/8)\n\nRemember that log(a^b) = b * log(a), so bring the -t/8 in front:\nlog(0.01) = -t/8 * log(2.2)\n\nNow just solve algebraically:\n-t/8 = log(0.01) / log(2.2)\nt = -8 * log(0.01) / log(2.2)\n\nlog(0.01) = -2\nlog(2.2) ≈ 0.342422681\n\nt = (-8 * -2) / 0.0.342422681\nt = 16 / 0.342422681\n\nTherefore:\nt ≈ 46.7258768\n\nAs a double check, you can plug it in your calculator:\n2.2^(-46.7258768/8) = 0.01
807860
Yes, though don't forget the y on the second log:\n\nlogb x^2/3 + logb y^3/4\n\nAlso, if you would have had:\nlogb x^2/3 / y^3/4 it would equal logb x^2/3 - logb y^3/4\n\nlog rules are just like exponent rules: multiplication = addition and division = subtraction
962855
should be e^u*du/dx\n\nd[e^(x^2)] = e(x^2)*2*x
920193
e^(xlnx )=x^x\nderivative is  1+ln x
158569
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
627488
you can change the base...\nln x = (log2 (x))(ln 2)