85 = 10*log(B/10^-16)  (What is B?)\n\nI took a math test tonight and completely blanked on the question, could not figure it out.\n\nEventually I came up with an answer of 1.70x10^-4, I do not know if this is right or wrong.\n\nIf anyone knows how to work the problem, please let me know cause it is driving me crazy.
132407
Since this is college algebra, it should be tough, but I'm having reading errors on the problem.\n\nCan't you just simply it down?\n\n8.5 = log (B/10^-16)\n\n10^8.5 = B/10^-16\n\nB = 10^-7.5
385162
Remember that with logarithms, exponents within the log become the coefficient outside the log.\n\n1.) logc (sqrt x)\n= logc (x^1/2)\n= (1/2) logc x\n= (1/2) * 1.2\n= 0.6\n\nIf numbers are being multiplied within the log, they can be split up to the sum of the log of the numbers.  It's a similar rule to that in problem 1, as you can imagine:\n\nlog (x^2) = log (x * x) = log x + log x = 2 log x\n\n2.) logc xy\n= logc x +logc y\n= 1.2 + 3.9\n= 5.1
457143
2xlog10=log9037980\n2x=6.9561\nx=6.9561/2\n=3.4781
17429
its simple ... 80*70*60 ... 10*0 = 0 !!\n\nright isnt it ??? :D
382784
1) The definition of a logarithm says : if a^X = N, then X = log[a]N.\nTurn this around to say : if X = log[a]N, then a^X = N.\nNow you can see what to do if we let X = log[2]64.\nWe get : 2^X = 64 = 2^6.\nTherefore, X = 6, that is, log[2]64 = 6.\n\n2) Same with this one. Let X = log[2]sqrt(2).\nThus, 2^X = sqrt(2) = 2^(1 / 2).\nTherefore, X = 1 / 2.\n\n3) 5^(-3) = 1 / 125\nUse the definition again. If a^X = N, then X = log[a]N.\nHere we have : a = 5, X = -3 and N = 1 / 125.\nThus, -3 = log[5](1 / 125)\n\n4) 10^(0.3010) = 2\nIn the same way, let a = 10, X = 0.3010 and N = 2.\nThus, 0.3010 = log[10]2.\n\n5) t = log[4]7\nDefinition again : 4^t = 7.\n\n6) log[10]7 = 0.845\nDefinition again : 10^(0.845) = 7.\n\n7) Let X = log[64]4\nBy the definiton : 64^X = 4.\nOr, (2^6)^X = 4.\nOr, 2^(6X) = 2^2.\nTherefore, 6X = 2, or, X = 1 / 3.
202210
Hiya, right here goes - \n\nyou want to find log10 . where log is to the base 2.\n\nfrom the prvious lecture notes you can see that the starting point fot the discrete logarithm is to fix a primative element which is dependent on which base you are working to. By definition this is an element g such that\n\ng^i ≠ 1 (mod p) for 1<i<p-2  ( and p>10 for it to work)\n\nin the example g=2 and p=11. p has been chose as 11 because\n\n 2^i ≠ 1 (mod p)  for 1<i<9\n\nhe point of the logarithm is then to find i such that 2^i ≡ 10 (mod p) trial and error gives that 2^5 ≡ 10(mod 11)\n\nso log10=5
539869
Can you write out in words what the above equation is supposed to be?  It looks like either a base of 2 or a base of 1/32 and a log of the other of those two.  Remember you can write log base a of b as log(a)/log(b) and that log(a^n) = n*log(a).  By those two properties, if the above is log base 2 of 1/32, then it is the same as log(1/32)/log(2) = log(1/(2^5))/log(2) = -5*log(2)/log(2) = -5.  Hope this helps, but clarify if this isn't what you were asking.
521878
The 'good news' is that this problem does not involve calculus, only number theory (or counting).  The 'bad news' is that very few students 'get it' when it comes to bases other than 10 (after all, we use base 10 numbers daily, so we tend not to think about how they work!).\n\nI think the best way to deal with this subject is to take a minute to think about "regular numbers" like 276.  Say it out loud, "Two hundred seventy six."  If you literally started with THREE numbers -- a 200, a 70, and a 6 -- and added them together, you would end up with 276.  Right?\n\nWell, picking a number apart by its digits (like the '2', the '7' and the '6' above) is exactly what you do in any base.  Except, instead of 'hundreds', 'tens' and 'ones', you have to find "powers of the new base".\n\nPowers of 10, of course, are easy.  We call them names like 'thousand' (10^3) or 'million' (10^6).  But what are the powers of, say, 8?\n\nLet's list them:\n8 raised to the 0 power?  That's 1.  Any base to the zero power is 1.  So that's nice.\n\n8 raised to the 1 power?  That's 8.  Any base raised to the one power is 'itself.'\n\n8 raised to the 2 power?  That's 64.  Any base raised to the two power is 'itself times itself.'\n\n8 raised to the 3 power?  That's 512.  As you might have guessed, that's just 8*8*8.\n\nAnd so on... we only needed those powers for your problem.  Because the given number (3175) only has four digits!\n\nLet's take each digit, starting with the 3 and working our way toward the right.  We'll multiply each digit times the power of 8 (the given base) that the digit 'stands for'.  At the end, we'll add everything up!\n\n3 times 8^3 = 3 * 512 = 1536\n1 times 8^2 = 1 * 64   = 64\n7 times 8^1 = 7 * 8     = 56\n5 times 8^0 = 5 * 1     = 5\n\nAdd them all up, now.  = 1661 (in base 10)\n\nWe've just 'converted' to base 10 this way.  Does it seem strange that the 'same' number can be written different ways using different bases?  3175 (base 8) = 1661 (base 10)\n\nThis number could be represented a zillion different ways if you keep changing the base!\n\nBut the problem asks for you to convert it to base 830.  In fact, it's unclear whether we should now convert "1661" (the n we found) or "1662" ("the integer immediately after n").\n\nLet's convert 1662 for fun, into base 830.\n\nLucky for us, the powers of 830 get BIG quickly, and we don't need more than the first couple of powers:\n\n830 ^ 0 =    1 (as always)\n830 ^ 1 = 830\n830 ^ 2 = ___ (way too BIG)\n\nAll we need to do is ask "How many times will these powers of our new base go into 1662?"  And then repeat that question with the remainder.\n\n1662 divided by 830 is "a little over 2".  (Because 830 * 2 is only 1660.)  So, the first digit of the answer is '2'.  \n\nWhat is the remainder?  Subtract:  1662 - 1660 = 2.  The remainder happens to be two.\n\nHow many times will the next-lower power of 830 go into the remainder?  Lucky for us, the next-lower power of 830 is "the 0 power", so we're thinking about 1.  How many times does 1 go into two?  Exactly twice!\n\nSo the final digit of our answer is ALSO 2.\n\nThe answer is (put the digits together): 22!\n\nNeat, huh?
434856
Question 1 : 4^x - 2 * 5^(2x) < 10^x\n\nFactorise :\n2^(2x) - 2 * 5^(2x) < 2^x * 5^x\n\nDivide through by 2^(2x) :\n1 - 2 * (5/2)^(2x) < (5/2)^x\n\nLet y = (5/2)^x\n\nTherefore, 1 - 2y^2 < y\nor, 2y^2 + y - 1 > 0\nor, (y + 1)(2y - 1) > 0\nThus, y < -1 or y > 1/2\n\nThat is, (5/2)^x < - 1 or (5/2)^x > 1/2\n\nBut (5/2)^x is always > 0, so the only\nsolution occurs when (5/2)^x > 1/2.\n\nTake the reciprocal of both sides :\n(2/5)^x < 2\n\nTake logarithm of both sides :\nx(log2 - log5) < log2\n\nTherefore, x < log2 / (log2 - log5)\nor x < -0.7565 approximately.\n----------------------------------------------------------------------\nQuestion 2 : 7 * 3^(x + 1) - 5^(x + 2) = 3^(x + 4) - 5^(x + 3)\n\nDivide through by 3^(x + 1) :\n7 - 5 * (5/3)^(x + 1) = 27 - 25 * (5/3)^(x + 1)\n\nRearrange :\n20 * (5/3)^(x + 1) = 20\n\nTherefore, (5/3)^(x + 1) = 1 = (5/3)^0\nThus, x + 1 = 0, so, x = -1\n----------------------------------------------------------------------\nQuestion 3 : 6 * 9^(1/x) - 13 * 6^(1/x) + 6 * 4^(1/x) = 0\n\nFactorise for ease of viewing :\n6 * 3^(2/x) -13 * 2^(1/x) * 3^(1/x) + 6 * 2^(2/x) = 0\n\nDivide through by 6^(1/x), which equals 2^(1/x) * 3^(1/x) :\n6 * (3/2)^(1/x) - 13 + 6 * (2/3)^(1/x) = 0\n\nThe latter term, (2/3)^(1/x) = 1 / [ (3/2)^(1/x) ]\n\nSo now we have : 6 * (3/2)^(1/x) - 13 + 6 / [ (3/2)^(1/x) ] = 0\n\nLet y = (3/2)^(1/x)\n\nTherefore, 6y - 13 + 6 / y = 0\n\nMultiply through by y :\n6y^2 - 13y + 6 = 0\nor, (3y - 2)(2y - 3) = 0\nor, y = 2/3 or 3/2\n\nThat is, (3/2)^(1/x) = 2/3 or (3/2)^(1/x) = 3/2\n\nFirst, trying (3/2)^(1/x) = 2/3 = 1 / (3/2) = (3/2)^(-1)\nTherefore, 1 / x = -1, so x = -1.\n\nSecond, trying (3/2)^(1/x) = 3/2 = (3/2)^1\nTherefore, 1 / x = 1, so x = 1.\n\nSo there are 2 solutions, x = -1 or 1.
95509
Multiply the numbers 1.25 and 8\n1.25*8 = 10\nadd the exponents on the 10's\nneg.3 + neg.5 = -3 + -5 = -8\nThus, we have\n10 times 10^(-8)\n10 is not in scientific notation\n10 = 1 times 10^1\nThus, the final answer is\n1 times 10^(-7)
347511
125 = log(x^2/1.5) \n125 = 2/1.5 logx  [using logx^m = m logx]\n(125*1.5)/2 = logx\n93.75 = logx\nx = antilog (93.75) OR x = e^93.75
294923
Yes you can use any base logarithms to multiply and divide positive numbers.\n\nBy simplify, I presume you mean 'calculate the answer'.\n\n9.813617548 + 9.760194383 - (-7.398778275) = 26.97259021\n\ne^26.97259021 = 5.17663 x 10^11\n\n(I cheated - I used Excel)\n\nI guess you knew that to multiply, you add the logarithms and to divide, you subtract the logarithm.
209170
>> log (x+3) + log (2-x) = 1\n\nlog( a b ) = log( a ) + log ( b ) so\nlog (x + 3) + log (2 - x) = log( (x + 3)(2 - x) ) = 1\n\nlog(e) = 1 if using natural logarithms, log(10) = 1 if using base 10 logarithms\n\nFor the former you have (x + 3)(2 - x) = -(x^2) - x + 6 = e and for the latter you have -(x^2) - x + 6 = 10.  These are both quadratic equations which should be easy to solve.  For natural logarithms you have\n\n-(x^2) - x + 6 = e\nx^2 + x = 6 - e\nx^2 + x + (1/4) = (6 + (1/4) - e) = (25/4) - e\nx^2 + x + (1/4) = (x + (1/2))^2 = (25/4) - e\nx + (1/2) = +/- sqrt( (25/4) - e )\nx = -(1/2) +/- sqrt( (25/4) - e )\n\nAs a final check on suitability of these values, we see that numerically x = 1.379... or x = -2.379... and so both x + 3 and 2 - x are positive for both values of x and so you have no problem taking log( x + 3 ) and log (2 - x) with either value for x.\n\nWith base 10 logarithms, completing the square [i.e., solving the quadratic] would lead to complex roots:\n\n-(x^2) - x + 6 = 10\nx^2 + x = -4\nx^2 + x + (1/4) = -4 + (1/4) = -(15/4)\nx^2 + x + (1/4) = (x + (1/2))^2 = -(15/4)\nx + (1/2) = +/- (1/2)sqrt(15) i\nx = -(1/2) +/- (1/2)sqrt(15) i\n\nSo probably the natural logarithm was intended. :-)\n\n>> 2^x+1 =5^1-2x\n\nIf you mean (2^x) + 1 = (5^1) - 2x then you have\n(2^x) + 1 = 5 - 2x\n(2^x) = 4 - 2x\nand "by inspection" x = 1 is a solution.\n\nIf you graph both y = 2^x and y = 4 - 2x you find only one point of intersection; alternatively, the derivative of 2^x - 4 + 2x is strictly positive, so the function is strictly increasing and continuous and so has a single zero.\n\nEither way x = 1 is the only real solution to\n(2^x) + 1 = (5^1) - 2x\n\nHowever, if you meant\n2^(x + 1) = 5^(1 - 2x)\nthen you get (x + 1) log(2) = (1 - 2x)log(5)\n(log(2) x) + log(2) = log(5) - (2 log(5) x)\n(2 log(5) + log(2)) x = log(5) - log(2)\n(log(5 * 5 * 2)) x = log(5 / 2)\nx log(50) = log(5/2)\nx = log(5/2) / log(50)\n\n>> p = a/1+be^-kt \n\nIf you meant p = a / (1 + b(e^(-kt)))\nthen with all log's to base e you have\n\na = p (1 + b(e^(-kt)))\nb = ((a / p) - 1) (e^(kt))\ne = inverse log of ( log(b) - log((a / p) - 1) ) / (kt)\nk = - (1/t) log(((a / p) - 1) / b)\np = a / (1 + b(e^(-kt)))\nt = - (1/k) log(((a / p) - 1) / b)
208914
3^x = 10.4\n\nsince both sides are always positive, you can take the log of both sides...\n\nlog(3^x) = log(10.4)\nx*log(3) = log(10.4)\nx = log(10.4)/log(3)\n\nYou can use a calculator to figure out the log of 10.4 and 3.\n\nSteve
256469
If you mean "to" 10%, the answer is:\n\n.64^n = .10\n\nn * (log .64) = log .10\n\nn = (log .10)/(log .64)\n\nn = -1/-.1938\n\nn = 5.16\n\nTherefore, 6 reductions would be needed.\n\n***EDIT***\nmathgirl826, after 1 reduction, there is 64% left.
349105
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)
466582
[ 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)