Does anyone understand logarithms?
921542
Logarithms are exponents, so they follow the rules of exponents.  For instance,\n\n(x^a)*(x^b) = x^(a + b)\n\nTherefore, \n\nlog a + log b = log(ab)\n\nAlso, every exponential equation has a logarithmic form, and vice versa.  So, assuming we're talking base 10,\n\nlog x = y\n\nmeans the same as\n\n10^y = x\n\nSo,\n\nlog a + log b = log (ab)\n\nIn your case:\n\nlog(x) + log(x + 15) = 2\n\nlog(x(x+15)) = 2\n\nIn exponential form, this becomes\n\nx(x + 15) = 10^2\n\nx^2 + 15x = 100\n\nx^2 + 15x - 100 = 0\n\n(x + 20)(x - 5) = 0\n\nx = -20 or x = 5\n\nSince you can't take the log of a negative number, we reject x = -20, so the answer is\n\nx = 5.\n\nDoes that help?
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)]
541208
it is a proving isn't it?\nright handside: [logbase10 1000]/logbase10 2\n=[logbase10 10^3]/[(logbase2 2 )/( logbase2 10)]\n=3[logbase10 10]x[logbase2 10]\n=3logbase2 10\n=logbase2 1000 , proved\n\n\n\nif its right do mark it best i'll come to know i have done correct or not.\nsorry here is the right answer\nanswer is 4\ni got it by\nlogbase2 1000=1ogbase10 1000 / logbase100 "x"\n3logbase2 10=31ogbase10 10 / logbase100 "x"\nlogbase2 10=1/logbase100 "x"\nlogbase100 10 / logbase100 2 =1/logbase100 "x"\n.5logbase100 100 / logbase100 2 =1/logbase100 "x"\n2logbase100 2=logbase100 "x"\nlogbase100 2^2=logbase100 "x"\n2^2="x"\n"x"=4
437222
Rules of logs:\n\n--log (a*b) = log a + log b\n--log (a/b) = log a - log b\n--log (a^b) = b log a\n\nSo log (y^2 / x^3) would break down as\n\nlog (y^2 / x^3) = log y^2 - log x^3\n=2 log y - 3 log x\n\nwhich is answer C in your list.
756495
log(base x) y=[log(base 10)y]/[log(base10)x]. I hope you can take up from here!
364460
log(x) in base a (for a>0) is simply ln(x)/ln(a). So if you are able to calculate natural logarithms, you can calculate logarithms in any base.\nA very simple technique to calculate natural logarithms is to use the identities:\nd/dx(ln(x))=1/x and ln(1) = 0. This implies that to calculate ln(y) for a given y>0, you can use finite approximations to the integral:\n\nintegral form 1 to y of 1/x = ln(y)\n\na very simple approximation is the following: for a given integer (big!) k of your choice, \n\nln(y) = sum for i=0 to k-1 of k/(k-i(y-1))\n\nHope this helps
881075
if these are decimal logarithms, i.e. log base 10, use the formula\n\nlog(ab) = log a + log b\n\nfor example,\n\nlog(417) = log (4.17 * 100) = log 4.17 +log100\n= .6201 + 2\n=2.6201\n\nlog0.417 = log 4.17 + log0.1 = .6201 -1 = -0.3799\n\nlog(.1) = log (10^(-1)) = -1
627488
you can change the base...\nln x = (log2 (x))(ln 2)
883916
Let log_2(x) denote the log of x with base 2.\n\nUse the identity that log_2(x) = ln(x) / ln(2). This identity holds because:\n\nRecall that, by definition, 2^(log_2(x)) = x.\n\nln(x) = ln(2^(log_2(x))) = (log_2(x)) (ln(2)), so dividing through by ln(2) shows the result.
887169
You're getting into logarithm.  If X/D = R^Y, then Y = log(X/D), where log is the logarithm in base R.
1075997
Since you use log rather than ln, I assuming base 10 (common) logs.\n\nLogrithms are exponents!!!!!\nThey are the exponent you hang on the base to get the number in question.  That is,  saying   log x = y is the same as saying 10^y = x.\n\nSo log x = -2 is the same as saying 10^(-2) = x  or x = 1/100.
496208
log(x) - log(1 + (x^2 + 1)^(1/2))\n\nWhere log is the natural logarithm, of course.
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
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
106841
Try the following resources:\n\n1. http://www.krysstal.com/logarithms.html\n2. http://www.physics.uoguelph.ca/tutorials/LOG/\n1. http://en.wikipedia.org/wiki/Logarithms
1132233
this question will require calcs using logarithm.\nhence will take time
867514
[ 1/4 LOG (16) ] - [ 2 LOG (6) ] + [ LOG (3) ]\n\n[ LOG ((16)^(1/4)) ] - [ LOG ((6)^2) ] + [ LOG (3) ]\n\n[ LOG (2) ] - [ LOG (36) ] + [ LOG (3) ]\n\n...........................the first two LOGS have the same base of 10, \n...........................so when subtracting LOGS we divide.\n\n[ LOG (2/36) ] + [ LOG (3) ]\n\n[ LOG ((2divided by 2) / (36 divided by 2)) ] + [ LOG (3) ]\n\n[ LOG (1/18) ] + [ LOG (3) ] \n\n...........................the above two LOGS have the same base of 10, \n...........................so when adding LOGS we multiply.\n\nLOG ((1/18) times (3))\n\nLOG ((1/18) times (3/1))\n\nLOG (3/18)\n\n..................................then reduce the fraction by dividing by 3/3.\n\nLOG ((3 divided by 3) / (18 divided by 3))\n\nLOG (1/6)   \n\nLOG ((6)^(-1))\n\n..........an exponent can be moved and multiplied to the front of the LOG.\n\n(-1) times [ LOG (6) ]\n\n - LOG (6)\n\n\nTherefore, \n\n[ 1/4 LOG (16) ] - [ 2 LOG (6) ] + [ LOG (3) ] =     - LOG (6)