Need help in logarithms?
916994
(log x)^3 = log (x^9)\n(log x)^3 = 9 log x\n(log x)^2 = 9\nlog x = +/- 3\nx = 10^3 = 1000  or x = 10^-3 = .001
574891
im pretty sure that you will have to use the parts method to solve it. but the answer is\n\nx(log(1-ie^(ix))-log(1+ie^(ix))+i(Li(-ie^(ix)-Li(ie^(ix))\n\ni=imaginary unit\nlog=natural logarithm\nLi=polylogarithm\n\nyou can check my answer to see if its right because i also thought that it was wrong. i dont know how the imaginary units could have shown up but o well.
368201
(1/2)logx = log(x^(1/2)) = log(sqrt(x))\n\nso for exemple 1/2log(9)=log(sqrt(9)) = log(3).\n\nsometimes it's useful, sometimes it's more useful to write the log of a sqrt as 1/2 the log of the argument.
398683
Use log you will find\n\nor put\n\nx=3.597285024\n\nIts done
899135
1. Move the log part to the other side to make it positive, and move the 5 to the other side to get -2 = log(x+2)\nThe base of log is 10, so this means basically 10^-2 = x+2\n\n10^-2 is 1/100 or 0.01 = x + 2\n\n-1.99 = x\n\n2. 2e^(2x) + 5e^x - 3 = 0\n\nFactor by FOIL:  (2e^x - 1)(e^x + 3) = 0\n\nSo either e^x = 1/2 or e^x = -3 (which is impossible)\n\nYou can do ln (natural log) of both sides to "cancel out" the e:\n\nln(e^x) = ln (1/2) becomes x = ln(1/2)\n\n3. logx - 1/2log(x+4) = 1\n\nMinus becomes divide and the 1/2 becomes an exponent when combining logs.  So the left side simplifies to be:\n\nlog(x/(x+4)^(1/2))   Remember, 1/2 as an exponent means square root so this is\n\nlog(x/ sqrt(x+4)) = 1\n\nAgain, log has base 10 so this is really 10^1 = x/sqrt(x+4)\n\nSquare both sides: 100 = x^2/(x+4)\n\nCross multiply:  100(x+4) = x^2\n\n100x + 400 = x^2\n\nx^2 - 100x - 400 = 0  \n\nUse quadratic formula (be sure to eliminate negative answer(s) because you can't find the log of a negative number)
894472
Use a calculator.  Enter [2] [log] for the first one.
808003
Do you know your log identities?  That would help you here a lot!\n\nUse these log identities to solve your problems:\n\n. log(a) + log(b) = log(ab)\n. log(a) - log(b) = log(a/b)\n. c*log(a) = log(a^c)\n\nSo, for problem 1 (I'm assuming the "a" means that "a" is your base) we get the following:\n\nlog a(y+5)+2log a (x+1) =\nloga(y+5) + loga((x+1)²) =\nloga((y+5)(x+1)²)  \n\nProblem 2:\n\nIn(x+5)-In(3)=In(x-3)\nln((x+5)/3) = ln(x-3)\n(x+5) / 3 = x-3\nx + 5 = 3x - 9\n2x = 14\nx = 7\n\nProblem 3:\n\n3^x+5=9^x\n\nIs 3 raised to the power of x only, or to the power of x+5?  It's hard to tell, and it does matter to the solution.\n\n~ ♥ ~
911082
logx+log(x+48)=2\n=>log{x(x+48)}=2\n=>log(x^2+48x)=2\n=>x^2+48x=10^2\n=>x^2+48x-100=0\n=>x^2+50x-2x-100=0\n=>x(x+50)-2(x+50)=0\n=>(x+50)(x-2)=0\nhence,\nx=-50 or x=2
746085
taking log on both sides\nlny=(x-5)ln2\n1/y(dy/dx)=ln2\ndy/dx=yln2=2^(x-5)ln2
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
978473
log(10) = 1, so \n3 = 3log(10) = log(10^3) = log(1000)\n\n2log(2) = log(2^2) = log(4)\n\n3 - 2log(2)\n= log(1000) - log(4)\n= log(1000/4)\n= log(250)\n\nI used the following properties:\n1) log(x^n) = n log(x)\n2) log(a) - log(b) = log(a/b)
743256
First 4*log[x-2] = log[x-2]^4. \nThen -log[5] = log[1/5]\n\nThe sum of logs is the log of the product of the arguments and so this gives\n\n(1/2)*log[(x-2)^4 * (x+3) / 5]\n\nFinally the (1/2) becomes an exponent in the argument:\n\nlog[((x-2)^4 * (x+3) / 5)^(1/2)]
783096
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
898977
log(2x+1)-(1/3)log(x+1)\n=log(2x+1) - log(  (x+1)^(1/3)  )\n=log( (2x+1) /   (x+1)^(1/3)  )
545001
1. 6^(-.2x)=10\n(-.2x)ln(6)=ln(10)\nx=-5*ln(6)/ln(10)\n\n2. 7log_5(x)=18\nlog_5(x)=18/7\nx=5^(18/7)
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
754734
1) The definition of a logarithm says : if a^X = N, then X = log[a]N.\nTurn this around to say : if X = log[a]N, then a^X = N.\nNow you can see what to do if we let X = log[2]64.\nWe get : 2^X = 64 = 2^6.\nTherefore, X = 6, that is, log[2]64 = 6.\n\n2) Same with this one. Let X = log[2]sqrt(2).\nThus, 2^X = sqrt(2) = 2^(1 / 2).\nTherefore, X = 1 / 2.\n\n3) 5^(-3) = 1 / 125\nUse the definition again. If a^X = N, then X = log[a]N.\nHere we have : a = 5, X = -3 and N = 1 / 125.\nThus, -3 = log[5](1 / 125)\n\n4) 10^(0.3010) = 2\nIn the same way, let a = 10, X = 0.3010 and N = 2.\nThus, 0.3010 = log[10]2.\n\n5) t = log[4]7\nDefinition again : 4^t = 7.\n\n6) log[10]7 = 0.845\nDefinition again : 10^(0.845) = 7.\n\n7) Let X = log[64]4\nBy the definiton : 64^X = 4.\nOr, (2^6)^X = 4.\nOr, 2^(6X) = 2^2.\nTherefore, 6X = 2, or, X = 1 / 3.