What is the answer to log base 2 1000 = logbase10(1000)/log base 10^2?
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
754734
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.
914213
OK. It looks like you took the log base 10 of 9 (which is .9542) and multiplied it by 1/81 (which is .01234) to get your .01178. Thus, you found\n(1/81)*Log(9).\n\nTo get log base 9 of 1/81, i.e.\nlog_9 (1/81)\nyou have to find the exponent x where\n9^x=1/81.\nThis exponent will be -2 since\n9^(-2)=1/9^2=1/81.\nNo calculator is needed here!\n\nTo find the result on a calculator, you have to find\nLog(1/81) (which is -1.908485)\nand divide it by\nLog(9) (which is .9542425)\nto get -2.
144204
-10\n\nand to prove it -10 x -10 x -10 = -1000
1187309
I think I did it.  If y=1/log(x), then a few manipulations gave me\n\nx=10^(1/y).\n\nI'm assuming by log you mean log to the base 10.  I suppose mathematicians will start fussing about what happens at y=0, but a physicist like me doesn't worry that much about infinities ...
789586
Giving you a similar analogy\n\n10^2 = 100\n\nso, Log 100 to base 10 = 2\n\nSimilarly, if 2^x=5\n\nthen Log 5 to base 2 = X\n\nand that is the answer\n\nTo express it nemerically, we need to do this way\n\n2^x = 5\n\nLog 2^x = Log 5 (taking logariths with base 10)\nor x Log 2 = Log 5\nor, x = Log 5/Log2 = 0.698/0.0.30102 = 2.321928095
445906
10 times 10 times 10 1000\nyou could use a calculator alot easier
790275
log(9)3\n=\nlog(9)9^1/2\n=\n1/2*log(9)9\n=\n1/2\n[as log(m)m=1]
449358
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
826204
Do you mean \n\nlog (base n) (x) = 1/(log (base x) (n))\n\nWell let log (base n) (x) = a\nTherefore x = n^a\n\nTake logs (base x)\nSo 1 = log (base x) (n^a)\n= a * log (base x) (n)\nBut a = log (base n) (x)\nTherefore log (base n) (x) * log (base x) (n) = 1\nie log (base n) (x) = 1/(log (base x) (n)) ............. QED
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.
437664
The secret is to understand some of the rules of logarithms and adjust translate the numbers to make your lob easier.\n\nThe rules:\nlog(a*b) = log(a) + log(b)\nlog(a/b) = log(a)- log(b)\nlog(a^n)=nlog(a)\n\nIf I understand your problem:\n\nx= [(86400)*sqrt(365.25)]/[ (.079)^11]\n\nso log(x) = log [.....]\n\n     = log(86400)+log(356.25^.5) - log(0.79^11)\n     = log(86400)+.5*log(356.25) - 11*log(0.79)\n\nuse log tables to get the logs and add them up to get the log of x.  Use the tables again to find x.\n\nYou can make other approximations, i.e., 356.25 is close to 19^2 (=361)\nand 86400 is close to 300^2 (=90000) and continue from there.  \n\nHope that helps.
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.
82029
it's a simple work: I'll give you a suggestion, but you should try to do it by yourself.\n\ny intercept: you find it by setting x=0 so you get\n\ny= log(base2)(2) + 3\n\nThe logarithm is simple to calculate ---> get an integer answer.\n\n---\n\nx intercept: you must set y=0 this time.\nYou get the logarithmic equation\nlog(base2)(x+2) = 3 ---> to solve for x, let each member be the exponent of a power with base 2:\n\n2^(logbase2(x+2)) = 2^3\n\n---> use definition of logarithm (why did we choose a power with base 2?) and get to an integer solution.\n\nGood work.
140923
The other property of logarithms is that Log_x A - Log_x B = Log_x (A/B)\n\n\nso you can combine this rule with the other one i told you about in your other question...\n\nthe only thing is that u are working in reverse order\n\n\nLn 2y/x= Ln 2+ Ln y - Ln x\n\n\nThese are properties, they should be tought to you as principals, without them you cant solve logs:D hope this helps ;) if you think it does pls gimme 10 pts:P
694278
Log(x + 2) + Log(x - 1) = Log(70)\nAs I explained before,\nLog (a0 + Log (b) = Log(ab)\nSo,\nLog(x + 2) + Log(x - 1) = Log [(x + 2)(x - 1)]\nLog [(x + 2)(x - 1)] = Log(70)\n(x+2)(x-1) = 70\nx^2 - x + 2x - 2 = 70\nx^2 + x = 72 \nx^2 + x = 72\nx^2 + x - 72 = 0\nx^2 - 8x + 9x - 72 = 0\nx(x - 8) + 9(x - 8) = 0\n(x + 9)(x - 8) = 0\nNow, either x + 9 = 0 or x - 8 = 0\nIf we put x + 9 = 0, we get x = -9\nIf we put x - 8 = 0, we get x = 8\nBut when we substitute x = -9 in the original problem we get Log (-7) + Log (-10) = Log(70)\nBut Log of negative numbers if undefined.\nSo, x = 8 is the only solution.
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.