Help solving logarithm?
812732
2^(4x) - 4(2^(2x)) - 12 = 0\n\nFirst, note that 2^(4x) is equal to (2^(2x))^2\n\n(2^(2x))^2 - 4(2^(2x)) - 12 = 0\n\nThen, solve it like a quadratic.  Let y = 2^2x and you get:\n\ny^2 - 4y - 12 = 0\n(y - 6) (y + 2) = 0\n\ny = 6\ny = -2\n\nSubstitute y = 2^(2x) back to get\n\n2^(2x) = 6\n2^(2x) = -2\n\nYou can throw away the second equation because you can neve get a negative result from 2^(2x).  So we just deal with the first one.\n\n2^(2x) = 6\nConvert to logarithmic form\n\nlog[base 2](6) = 2x\nx = log[base 2](6) / 2\n\nThat can be simplified even more, but I'll leave that assignment up to you.
1093339
No one on here is going to actually DO your homework for you.. but here is a place for you the get help -\n\nhttp://www.themathpage.com/aPreCalc/logarithms.htm\n\nhttp://en.wikipedia.org/wiki/Logarithm\n\nhttp://physics.mtsu.edu/~wmr/log_2.htm
387252
x = log(7)19\n\n7^x = 19\n\nx = (log19)/(log7)\nx = about 1.513
437664
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.
300591
First, use the identities for logarithms to simplify this.  When you do, you get ln(30/x) = ln30 - ln x.  Now, divide each of those terms by ln2 and you get ln30/ln2 - lnx/ln2.  Now, take the limit as x->0.  As x goes to zero from the right (the left hand limit does not exist), ln goes to minus infinity.
895627
log {x^8} = 8 log x\nlog {x^2} = 2 log x\n\nplug and chug answer is 4
1035877
The natural logarithm (over the real numbers) is defined only for positive numbers, so \n\nx-4>0\n\nin other words\n\nx>4.
808711
Welcome!!\n\nHere is the solution:\n\nlog(x+1) = log3 - log(2x-1)\nlog(x+1) + log(2x-1) = log3\nlog[(x+1)(2x-1)] = log 3\n\ntherefore,\n\n(x+1)(2x-1) = 3\n2x^2 - x + 2x -1 = 3\n2x^2 + x - 4 = 0\n\nyou need to use the quadratic formula to solve this:\n\nx = {-1 ± sqrt[1 - 4(2)(-4)] } / 2*2\nx =[ -1 ± sqrt(33)] / 4
878117
depending on where your other bracket (]) is:\nlog[4,000(1.0042)^12t - 10,000(1.0063)^12t]= log1 is true\nwhen t=-164.08 or -36.6146
496208
log(x) - log(1 + (x^2 + 1)^(1/2))\n\nWhere log is the natural logarithm, of course.
533385
Please believe TheoDorkx. He's the only one who understands logarithms.\n\nAnother way to solve logarithm problems is directly change from the logarithmic to the exponential form.\n\nln(2x) = 2 means that 2 is the power to which you raise the base of the logarithm, e, in order to get the result 2x.\n\nSo the equivalent exponential form is\n\n2x=e^2\n\ntherefore\n\nx=(1/2)e^2\n\nThis is the same thing TheoDorkx got with his approach. I offer it as a slightly different way to think about the problem.
651623
Try to use logarithmic tables.
1027143
log 2 x^2+1-log2^x\nlog 2 + log X^ + log 1 - (log 2^ + log X)\n=log 2 + A log X + log 1 - A log 2 - log X\n=log 2 + log 1- log X + A log (X - 2)\n=log (2 x 1) - log x + A log (x -2)\n=log 2/x + A log (x-2)\n=log 2/x + log (x-2)^\n= log (2/x X (x-2)^) = log 2/x(x-2)^
586612
remember your natural logs\n625^x = 1/125\nxln625 = ln(1/125)\nx = ln(1/125)/ln625 = -3/4
147502
log x goes to negative infinity\nbut with the squared it becomes positive.\nSo the limit is +infinity
313189
log2^3x = log64\n2^3x = 64\n2^3x = 2^6\n3x = 6\nx = 2
729730
u cant split a log like that! ie ln(ab) = lna +lnb\n\nu should use the f'(x) f(x)^n = f(x)^(n+1)/ (n+1)