log(2x+1)-Log(5)=log(x)?
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
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
978189
e^(ln5)=x   .  Now express this in log form by thinking of ln(5)\n                       as just an exponent.\n\n   lnx=ln5   so obviously\n\n   x=5.
746085
taking log on both sides\nlny=(x-5)ln2\n1/y(dy/dx)=ln2\ndy/dx=yln2=2^(x-5)ln2
815091
First of all, we apply the power rule:  a*log(x) = log(x)^a\n\n     2*log(a(x+1)) = log(a(x+1))^2\n\nNext, we know that the sum of two log expressions can be written as a single product: log(a) + log(b) = log(a*b)\n\n     log(a(y+5)) + log(a(x+1))^2 = log[a(y+5)*(a(x+1))^2]\n\nYou may try some algebraic simplification, but I think that these concepts are the ones that are being tested for this problem.
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)
754397
ln(x+3) - ln(x) = ln[(x+3)/3]\n\n2ln(2) = ln(2²) = ln4\n\nSo, if ln[(x+3)/x)] = ln4, then\n\n(x+3)/x = 4, or\n\nx+3 = 4x, or \n\nx = 3x, or x = 1\n\nI used the properties of logs to help me solve this.  I think you tried to make it too hard on yourself (which is very easy to do when dealing with the hated logarithms!).  Also, you brought e into it when you didn't really need to.  Since we had ln(x) = ln(y), essentially, then from there we can conclude that x=y without bringing e into play.  Anyhoo, here's the log properties I used in this solution.  (I use the word log but it applies to ln as well.):\n\nlog(a) - log(b) = log(a/b)\n\nrlog(a) = log(a^r)\n\nHope that helped!\n\n~ ♥ ~
780572
First let's remember some rules of logs.\nThe log of product is the sum of the logs.  That is,\nlob(a*b) = log(a) + log(b)\nThe log of a quotitent is the difference of them.  That is, \nlog(a/b) = log(a) - log(b)\nAnd finally, if you have an exponent, you can bring it in front of the log.\nlog(a^n) = n*log(a)\nRemember that you can only take the log of positive numbers.\nln[ x^2 / [sqrt(x) (2+x)^2] ]\n= ln(x^2) - ln[ sqrt(x) (2+x)^2]\n= 2ln(x) - [ln(sqrt(x)) + ln(2+x)^2]\n= 2ln(x) - (1/2)ln(x) - 2ln(2+x)\nYou can combine 2ln(x) - (1/2)ln(x)\n2ln(x) - (1/2)ln(x) = (2 - 1/2)ln(x) = (3/2)ln(x)\nAnswer: (3/2)ln(x) - 2ln(2+x)\nNote: sqrt(x) = x^(1/2)
499879
I'm trying to guess what the problems you were asked to do were, because the information you gave in posing the question was incomplete.\n\nIf f(x) = log(x)  (for log of any base)\n\nthen\n\nf(1/b) = log(1/b) = log(b^(-1))=-log(b)=-f(b)\n\n\nlog(MN^2) = log(M) + 2 log(N)\n\nor\n\nlog((MN)^2) = log(M^2) + log (N^2) = 2log(M) + 2log(N)\n\nThese are applications of the laws of logarithms.
350639
7^(2x + 1) = 3^(x + 2)\nlog(7)(x + 2) = log(3)(2x + 1)\nIn the second step, you switched the x+2 and 2x+1\nLine 2 should be\nlog(7)(2x + 1) = log(3)(x + 2)\nContinue solving as you did before and you will get the correct answer.\n\nAlso, you wouldn't write ln9 just because it is not in simplest form\nln(9) = ln(3^2) = 2ln(3)\nSimilarly for ln(49) = 2ln(7)
329438
log_5(X) + log_5(x-1)=log_5(6)\nlog_5(x(x-1)) = log_5(6)\nx(x-1) = 6\nx^2 - x - 6 = 0\n(x-3)(x+2) = 0\nx = 3 or x = -2\nYou cannot take the log of a negative number, so -2 cannot be a solution.\nAnswer: x=3\nFor the second part, find T(30) and T(75)\nT(30) = 70 + 100e^(-0.04(30))\n= 70 + 100(.3012)\n= 70 + 30.12\n= 100.12\n\nT(75) = 70 + 100e^(-.04(75))\n= 70 + 100(.0498)\n= 70 + 4.98\n= 74.98
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~ ♥ ~
859504
(x-7)(x-5)\nso x= +7, or +5
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?
902324
Hello dear \n\n▪log ( x +3 ) + log (x) = 1 \n{ as you know if base is a so log (xy) = log x + log y }\n▪log x(x + 3) = log 10 \ncancel " log " \n▪x(x+3) = 10 \n▪x ² + 3x - 10 = 0\n▪(x + 5)(x - 2) = 0\n▪x = + 2 \n▪x = - 5  { cant accept it ,  because 0 < a = b   <=>  log a = log b }\n\nGood Luck.
760330
First let me assert that \n  log(base b)(a)=1/log(base a)(b) as follows:\n\n let log(base a)b=z\n   a^z=b\n   a=b^(1/z)  . Now sub back for z giving\n   a=b^(1/log(base a)(b))  and convert to log form of base b\n 1)   log(base b)(a)=1/log(base a)(b)  \n\n Now follow this: \n    let  log(base c)(x)=y\n        c^y=x   Now take log(base a) of both sides\n    ylog(base a)c=log(base a)x  and  sub back for y\n log(base a)x  = log(base c)x{log(base a)c} and\n use 1) to convert log(base a)c giving\n   2)  log(base a)x  = log(base c)x/{log(base c)a.\n\n     Lastly, into 2)  substitute X^3+1 for x,  4 for a and 10 for c\n\nand we have \n  log(base 4)(x^3+1)  = log(base 10)(x^3+1)/{log(base 10)4}
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)