lim x->0+ of (log(x))^2?
147502
log x goes to negative infinity\nbut with the squared it becomes positive.\nSo the limit is +infinity
745440
Okay, sorry if the format is kinda weird... By the way, the long line of "---------------" means I divided the line above it by the number below it, i.e.\n68\n---     = 68 divided by 4.\n4       \nThis is the real equation:\n\nLog(x+2)+Log(x-1)=Log(70)\nLog(x)+2Log+Log(x)-Log=70Log\nLog(x)+Log(x)+Log=70Log\n2Log(x)+Log=70Log\n2Log(x)=69Log\n--------------------\n         Log\n2x=69\n--------\n    2\n______\n|x=69/2|\n*^*^*^*^*\nI think that's the right answer. I've been working on that in my classes, too.  : [)
615960
As the graph for y=logx will approach but never reach x=0, the value for log0 is undefined. However, it does APPROACH negative infinity.
716399
log x-log(x-3)= -1\nlog(x/(x-3))=-1\nx/(x-3)=10^-1=.1\nx=.1(x-3)=.1x-.3\n.9x=-.3\nx=-1/3
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
634484
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
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)
903727
-1 / ( 20 ( ln( 10 + e ) * ( 10 + e )^(10 * x^2 ) ) )
850657
In the logarithm, there are several number of rules you have to remember.  One of them is by combining two or more logarithms into one.\n\nFor example,\nlog (a) b + log (a) c = log (a) bc\nlog (a) b - log (a) c = log (a) (b/c)\na is the base of the logarithm.\nb and c is any value more than 0.\nThis kind of combination can only be done if and only if the logarithms have the same base.\n\nThe other rule involved in solving your question is as following:\nc log (a) b = log (a) b^c\na is the base of the logarithm.\nb is any value more than 0.\nc is any real number.\n\nIn your case,\nlog (a) (y + 5) + 2 log (a) (x + 1)\n= log (a) (y + 5) + log (a) (x + 1)^2\n= log (a) (y + 5)(x + 1)^2\n\nIf you would like to multiply it, \nlog (a) (y + 5)(x + 1)^2\n= log (a) (y + 5)(x^2 + 2x + 1)\n= log (a) (x^2y + 2xy + y + 5x^2 + 10x +5)\n= log (a) (x^2y + 5x^2 + 2xy + y + 10x +5)\n\nHowever, this only further complicate the answer, therefore log (a) (y + 5)(x + 1)^2 would be the best answer.
550920
5^x=.125\nTaking log (base e) on both sides\n\n-> xlog5=log(0.125)\n-> x= log(0.125)/log5 = -1.291099302
982509
Let log2 5 = x\n\nThen 2^x must be 5\n\nTake log of both sides but use log10 (common log, on calculator)\n\nx log 2 = log 5\n\nDivide both sides by log 2
978128
log(base a) b = logb/loga.\n\nSo take the log of X and divide by the log of 1/3.  Then press the +/- button because you're being asked for the negative of that value.
1125670
Let x = (-2)^(-2)^(-2)^(-2)^(-2)^(-2)^...\n\n(-2)^(-2)^(-2)^(-2)^(-2)^(-2)^... = (-2)^(-2)^(-2)^(-2)^(-2)^(-2)^...\nx = (-2)^x\nln(x) = x * ln(-2)\n\nNot a real number, that's for certain.  I have to brush up on my complex logarithmic functions.... :-(
854009
lim as x approaches 0 from the positive of (ln x) is negative infinity.\nlim as x approaches 0 from the positive of (sqrt x) is 0.\n\nif the numerator goes to negative infinity and the denominator goes to 0, the limit is negative infinity. it might look like L'hospital's rule is required, but it is not.
935326
You have got to use expansion series!\n\nX->0+ [e-(1+x)^(1/x)]/x =e/2 \nlet x=0+h\n=>as x->0, h->o\n\nh->0 [e-(1+h)^(1/h)]/h\n\nlet y=(1+h)^(1/h)\n=>log y=(1/h)log(1+h)\n=>y=e^[(1/h)log(1+h)]\n\nexpansion of log(1+x)=x - (x^2)/2..................\n=>e^[(1/h)log(1+h)]= e^[(1/h)[h - (h^2)/2!............]\n\n=>e^[(1/h)[-h - (h^2)/2!............]=e^[1-h/2!........]\n\ntherefore,\n\nlet y =lim h->0 [e-(1+h)^(1/h)]/h=lim h->0 {[e- e^(1-h/2!...)]/h}\n\nexpansion of e^x=1+x+(x^2)/2!.................\n\n=>lim h->0 {[e- e^(1-h/2...)]/h}=lim h->0 {-e[e^(-h/2.....) -1]}/h\n\nexpansion of e^x=1+x+(x^2)/2!.................\n=>lim h->0{-e[1+(-h/2-(h^2)/3.....)+ ........-1]/h}\n\n=>lim h->0{-e[(-h/2-(h^2)/2...)+ [(-h/2-(h^2)/2...)^2]/2!+.....]}/h\n\n=>lim h->0{-e(-1/2 -..(terms containing highter powers of h))}\n\n=>e/2\n\n\nP.S.  By the way L' Hospitals rule would fail here as limit is not of form 0/0 or ~/~. It is of form a/0 where a is constant.
778101
x^x = e^{xln(x)} \nnow, the lim x->0 x ln(x) = 0,\nso lim x->0 x^x = e^0 =1\n`\n`
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.