suppose y>x\nlogy>logx\nlog(y^5)-log(y^4)>\nlog(x^5)-log(x^4)\nrearranged\n-log(y^4)+log(x^4)>\nlog(x^5)-log(y^5)\n\nlog(x^4/y^4)>\nlog(x^5/y^5)\n\n4log(x/y)>\n5log(x/y)\n\nthus 4>5
210062
pablodiablo is right, you can't treat an inequality like an equal sign when dividing or multiplying both sides by a negative number...swap the sign when doing so.  if you look at the graph of a logarithmic function, if 0<n<1, then log(n) is a negative number, and since y>x, x/y is less than one.  also, x & y must both be the same sign (positive or negative), otherwise x/y will be negative, and log of a negative number is undefined.
422309
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
62442
[x^2/4][2(log(x) - 1) log(x) + 1]
429792
Remember that log x - log y =log (x/y) to rewrite that equation. Then, using base 10, you have the equation 10=(x+4/ x+3). Solve.
375872
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.  : [)
281840
log ( sqrt (x+ 1) *x^5)
410114
These require logarithmic differentiation.\n\nTake the ln of both sides.\n\n1. ln(y) = (x+2)*ln(x)\nDifferentiate with respect to x:\n\n(1/y)*dy/dx = (x+2)*1/x + ln(x)\n\nMultiply by y on both sides:\n\ndy/dx = y[(x+2)/x + ln(x)]\n\nReplace y:  dy/dx = x^(x+2)[(x+2)/x + ln(x)]\n\n2.  ln(y) = x*ln(x+5)\n\n(1/y)dy/dx = x*1/(x+5) + ln(x+5)\n\ndy/dx = y[x/(x+5) + ln(x+5)\n\ndy/dx = (x+5)^x[x/(x+5) + ln(x+5)]\n\n3.  Use the formula d/dx(a^u) = a^u*ln(a)*du/dx\n\nThen dy/dx = 3.14^x*ln(3.14)
59801
log2x=5\nx=2^5. You change the problem to exponent form. Your base is 2 you exponent is 5 and your soution is x.\nlog    solution= exponent    \n   base                     \nthis may be alittle confusing but I'm positive you'll get it.:) keep trying
196132
The secret is to understand some of the rules of logarithms and adjust translate the numbers to make your lob easier.\n\nThe rules:\nlog(a*b) = log(a) + log(b)\nlog(a/b) = log(a)- log(b)\nlog(a^n)=nlog(a)\n\nIf I understand your problem:\n\nx= [(86400)*sqrt(365.25)]/[ (.079)^11]\n\nso log(x) = log [.....]\n\n     = log(86400)+log(356.25^.5) - log(0.79^11)\n     = log(86400)+.5*log(356.25) - 11*log(0.79)\n\nuse log tables to get the logs and add them up to get the log of x.  Use the tables again to find x.\n\nYou can make other approximations, i.e., 356.25 is close to 19^2 (=361)\nand 86400 is close to 300^2 (=90000) and continue from there.  \n\nHope that helps.
663331
s = c(1 - d)^n\n(1 - d)^n = (s/c)\nlog(1 - d)(s/c) = n\n\nANS : n = log(1 - d)(s/c) or n = log(1 - d)s - log(1 - d)c\n\nWhereas (1 - d) is the base\n\nor you can type it like this\n\nn = (log(s/c))/(log(1 - d))\nor \nn = (log(s) - log(c))/(log(1 - d))
387879
2^(5x - 3) = 5^(x - 4) \n(5x-3)log2 = (x-4)log5\nx(5log2 - log5) = 3log2 - 4log5\n\nx = (3log2 - 4log5)/(5log2-log5)
62162
It would be nice if you could just flip, but...\n\nYou need to change your y to x and your x to y.\n\nFor example:\n\ny=3log(x+6), would become\n\nx=3log(y+6), solve for y \n\nx/3 = log(y+6)\n\n10^(x/3) = y+6\n\ny = 10^(x+3)-6\n\nUse the same process for the others.
95569
it simplifies to 2x^n=x^n+1 \n\ntake the log of both sides (log x)\n\nlogx^2n=logx^n+1\n\nnlogx^2=(n+1)logx \n\nn&#92;(n+1) =logx/logx^2\n\nn/(n+1) = 1/2\n\n2n=n+1\nn=1\n\nPlug n back into equation\n\nx^1 +x^1 = X^2\n\nx+x=x^2\n2x=X^2\nx^2-2x=0\nx(x-2) = 0\nx=0 or x=2
402804
OK, there's 2 possibilities:\nIf you meant:\nlog (16^1/4), then you need to know the property of logarithms where:\nlog a^b = b log a.\n\nSo:\nlog 16^(1/4) = 1/4 log 16\nlog 10^(.01) = .01 log 10\nlog 2^(1/8) = 1/8 log 2\n\nIf you meant (log 16)^1/4, then I have no idea how to do it by hand.\n\nEdit:\nFurther, you can also do things like:\nlog 16 = log 2^4 = 4 log 2\nso 1/4 log 16 = 1/4 * 4 log 2 = log 2.
451886
log3x - 3[logx - log(1+y)]\nwe know \nn log A = log [ A ^n]\nlogA - logB = log [A/B]\nso, going from right to left,\nlog 3x - 3[ log [ x/(1+y)]\nlog3x -  ( log [(x/(1+y)) ^3])\nlog [ 3x / (x/(1+y)) ^3) ]\nlog [ 3 * ( ( 1+y) ^3) / x^2  ]\n:)
403896
yes, it really is easier than you think:\na^(log_a(2/x) +log_a(x))=a^(log_a(2) -log_a(x) +log_a(x))\n  =a^(log_a(2))\n  =2\nnote that i used the property log(a/b)= log(a)- log(b) and a^log_a(x)=x
462950
When we need to ind the logarithms of number not of base ten, we can use log*6*x=logx/log6 so the answer is\n(log60)/(log6)-(log30)/(log6)