Stupid question I know but how would I solve this... log x ^ log x = 25?
819937
You haven't put brackets in, so the question looks ambiguous.\n\nIt could be [log(x)] ^ [log(x)], which is difficult,\nor it could be log[x ^ log(x)], which is easier.\n\nI'll take the easy way out.\n\nLet log[x ^ log(x)] = 25, where I'll assume log is to the base 10.\n\nBy one of the rules of logarithms, which states that\nlog(M ^ N) = N * log(M), we have :\n\nlog(x) * log(x) = 25\nor, [log(x)] ^ 2 = 25\n\nTaking the square root of both sides gives :\nlog(x) = ± 5\n\nAs I've assumed this is to base 10, this gives :\nx = 10 ^ (± 5)\nSo, x = 10 ^ 5 or 10 ^ (-5),\nwhich is 100,000 or 0.00001.
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.  : [)
895627
log {x^8} = 8 log x\nlog {x^2} = 2 log x\n\nplug and chug answer is 4
575720
take logbase6 of 135 = x\n\nlog 135/log 6 = x\n2.737 = x\n\nThen substitute this into original to get:\n2^(log135/log6) + 1.3^(log135/log6)-2 \n\nIt should prove that it equals 30, which proves that both statements are ture.  However, I am getting that it is not equal to 30...so I think the proof is impossible,  unless your formula is typed incorrectly.
157110
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.
427245
You can only take the log of numbers that are positive, so\n\nx^2 + 6x + 5 > 0\n\nSo (x+5)(x+1) > 0\n\nSo x would have to be greater than -1 and less than -5.
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
941613
log_a b = x is the number to which you need to raise a to get b:\na^x =b\n...........
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.
1046114
80=16*5\n\nlog(2^4*5)/log2\n=(log2^4+log5)/log2\n=(4 log2+log(10/2))/log2\n=(4log2+log10-log2)/log2\n=(3log2+1)/log2=3+(1/log2)
674167
Solve this by using logarithms\nlog((1/5)^2x) = log(625)\n2x*log(1/5) = log(625)  Product Rule\n2x = log(625)/log(1/5)\n2x = -4\nx = -2\n\ntosmith
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
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
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
324025
log 10000 = x\n10^x = 10000\nx = 4\n\n4
889767
log2 8 is log8/log2\nlog2 32 is log32/log2\n\nnow just plug it in a calculator!\nyou should get 8.\n\nif you can't use a calculator, set it up like this:\n2^x=8 and 2^y=32\nthen add x and y (3 and 5)
118410
...=(logA+logA)-(logB-logB-logB)+1/5logC=(logA^2-logB^3)+1/5logC\n= log(A^3/B^3)+1/5logC =log(A^3/B^3)+logC^1/5=log[(A^3*C^1/5)/B^3]\nNow if you know to transform C^1/5 (i don't really remember now)it becomes even simpler.