Is the only way to solve n = (log 0.05) / (log -0.6478) that has negative log numbers is only guessing and can
823338
That was my answer you quoted above, and when I encountered the "log of a negative number," I did <not> just make a willy-nilly guess, so in my opinion, your characterization was unwarranted.  (In fact, you had asked me to look at this problem in the first place.)\n\nWhen I removed the negative sign, I obtained an answer, and, if you looked at the "sources" section, you'd have seen that it was the right answer.\n\nIn light of your denigration of what you called my "guessing method," I'll add the following (which I felt I didn't have to research in light of the correct answer I obtained and checked) ...\n\nThe logarithms could just as easily have been natural or Napierian logs instead of common logs.  Instead of\nn = (log 0.05) / (log -0.6478), I could have written\nn = (ln 0.05) / (ln -0.6478).\n\nBut ln(-0.6478) = (ln -1) + (ln 0.6478)\nwhich begs the question, what is ln(-1)?\n\nTo answer this, we turn to Euler's famous identity,\n\ne^(i pi) + 1 = 0\ne^(i pi) = -1\n(i pi) ln e = ln(-1)\n\nBut since ln e = 1, we have\n\nln(-1) = i pi\n\nwhich is entirely in the imaginary portion of the complex plane, with no real component.\n\nTherefore, the real part of\n\nln(-0.6478) = (ln -1) + (ln 0.6478) = ln 0.6478\n\nConsidering what I'd done earlier with your question, including checking the series solution in the "sources" section, I thought I'd done enough.  Your characterization of my work is an insult.\n\nThe other answer given here -- a good one -- looks more general to me.  Instead of being based on the Euler identity that I used here, it's based on Euler's more general formula\n\ne^(ix) = cos x + i sin x\n\nNevertheless, I felt it necessary to defend what I'd done.
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.
15458
A negative number cannot have a square root.
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.
881529
fist domain\n\nthe number we search the log must be positive so x>-2 \n\ndomain -2<x<infinity \nfor infinity log (base3), infinite is infinite. for x=-2 we tend to x+2 =0 and log (x+2 ) = -infinite  Vertical asypmtote for x=-2 \n\nthe equation can be rewritten  h(x)  = log (base 3) (x+2/7^3)
812261
You are very close. If K = -0.032 is accurate enough, then use it.\n\npA = Ae^(19 * -0.032)\np = e^(19 * -0.032) ~ 0.544\n\nIf you leave K = ln(.5)/22 then\np ~ 0.5496
398683
Use log you will find\n\nor put\n\nx=3.597285024\n\nIts done
100388
This is due to a property of log which is that :\nlog(ab)=log(a)+log(b)\nRemark that this is true when well defined (for eg you can't write log(-1*-3)=log(-1)+log(-3) )\n\nSo here, you have log(x^2)=log(x)+log(x)\n\nTherefore, there are no diff. except that log(x^2) exists for all x in R&#92;{0} and 2log(x) is defined for all x strictly positive.
746085
taking log on both sides\nlny=(x-5)ln2\n1/y(dy/dx)=ln2\ndy/dx=yln2=2^(x-5)ln2
307451
Try\n\nhttp://www.purplemath.com/modules/negative.htm
586612
remember your natural logs\n625^x = 1/125\nxln625 = ln(1/125)\nx = ln(1/125)/ln625 = -3/4
708168
alog(x) = log(x^a)\n\n1/4log_b(16) = log_b(16^1/4) = log_b(2)\n\n2log_b(5) = log_b(5^2) = log_b(25)\n\nlog_b(7) doesn't change.\n\nlog(x) + log(y) = log(xy)\n\nand \n\nlog(x) - log(y) = log(x/y)\n\nTherefore\n\nlog_b(2)-log_b(25)+log_b(7) = log_b(14/25)
968809
The logarithm equation you presented is the change of base law.\n\nC can be any number, commonly 10 or e  as our calculators only work in bases of 10 or e and this law helps to solve log equations of other bases. \n\ne is the natural logarithm number (its like the concept of pi, e= approx. 2.71....)
470800
logx 1/a = (logx 1) - (logx a)\n= 0 - 3\n= -3
704178
I'm sure you must have been given these few principles of log operations:\n1. log a + log b = log ab\n2. log a  - log b = log (a/b)\n3. log a^p          = p*log a\n4. log 1 = 0\n5. log (subscript b) b = 1\n\nIn the first problem, if you apply rule 3 it becomes\n1/5(log(x+1)^3 + log(x-1)^2 - log 7), then apply rules 1 and 2 and you get\n1/5log(((x+1^3)*(x-1)^2)/7), a horrible looking expression and I don't know why anyone would ask you to touch it!\n\nIn the second one, log (subscript b) x*b\n= log x + log b [by rule 1]\n= log x + 1     [rule 5]\nTo deal with log sqrt 10, remember \nsqrt 10 = 10^(1/2) and use rule 3;\nalso 10 = 2*5 and use rule 1 in reverse, i.e. split log 10 into log 2 and log 5.\n\nI can't do much more without actually doing them all for you, and I don't think I should do that.\n\nIn  problem 3, take the log of each side and use rule 3, i.e.\nto solve  a^(x + b) = c, do\nlog a^(x + b) = log c\nand then (x + b)*log a = log c\n\nYou can';t do q.4 without knowing what the base of logs is.  I'd assume it's 10 and use rule 1 and solve\nlog x(x+3) = log 10\n\nIn problem 7, I think those 4's are all subscripts, in that case you are right and have used the rule I omitted: \n\nlog (subscript b) x  = (log x)/(log b)\n\nHowever, if you apply rule 3 to log 64, using 64 = 4^3, you'll simplify your answer a great deal.  Is it 3?
588312
-13 - (-10) + (-64) = -64\n\nThe Working : \n\n-13 - (-10) + (-64)\n=-13 + 10 - 64 = -67 \n\nNote:\nthe minus and minus in between 13 and 10 changed to a plus b'coz\n\n- and - = +\n\nBest of Luck !!
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.