log2 10 = 5 since 25 ≡ 10 (mod 11)?
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
641969
log 35 = log (7 * 5) = log 7 + log 5\n\nlog 8 = log (2^3) = 3 log 2 = 3 * 0.3010\n\nlog 5 is already given isn't it. \n\nand by the way you have written the value of log 7 wrongly. the value that you have given is of log 3.
911082
logx+log(x+48)=2\n=>log{x(x+48)}=2\n=>log(x^2+48x)=2\n=>x^2+48x=10^2\n=>x^2+48x-100=0\n=>x^2+50x-2x-100=0\n=>x(x+50)-2(x+50)=0\n=>(x+50)(x-2)=0\nhence,\nx=-50 or x=2
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.
916994
(log x)^3 = log (x^9)\n(log x)^3 = 9 log x\n(log x)^2 = 9\nlog x = +/- 3\nx = 10^3 = 1000  or x = 10^-3 = .001
771554
(x+5)(x-5)\n(x)^2-(5)^2\n=x^2-25
178809
Have you considered some sort of logarithmic compression?  for example, 10*log10(x+1)\nthen 10*log10(1+1) = 3\n10*log10(3+1) = 4\n10*log10(2+1) = 3.7\n10*log10(2.5+1) = 3.?\n10*log10(400 +1) = 26\n10*log10(500 +1) = 27\n\nother types are possible using the natural instead of the base 10 log.   there may be some rounding errors in my numbers, i wasn't using a calculator.\n\nI hope I didn't misunderstand your qustion.
978753
log 3x = log 5 + log (x-4)\nlog 3x = log 5(x-4)\n10 ^ (log 3x) = 10 ^ (log 5(x-4))\n3x = 5(x-4)\n3x = 5x - 20\n-2x = -20\nx = 10\nAnswer checks out.
765100
Have you studied logs yet?  You need to bring logs into this I think!\n\nRecall the log properties, as we will be using these:\nln(ac) = ln(a) + ln(c)\nln(a/c) = ln(a) = ln(c)\nln(a^r) = r*ln(a)\n\nSo you have y = -10*36^(1-2x) + 5\n\nFirst rewrite as this:  (5-y)/10 = 36^(1-2x)\n\nNow take the ln of both sides:\n\nln ((5-y)/10) = ln(36^(1-2x))\n\nNow use our log properties to get:\n\nln(5-y) - ln(10) = (1-2x) ln(36)\n\nDivide both sides by ln(36):\n\n( ln(5-y)-ln(10) ) / ln(36) = 1-2x\n\nHope this helped a bit.  I didn't solve everything for you, but hopefully this points you in the right direction.\n\nGood luck!
230715
We know that 2^10=1024.1024/11 leaves 1.Let x=2^10.Since 11 is a factor of (x-1) which itself is a factor of (x^n-1) (to prove this put x=1 in x^n-1).Now 2^(10*n)-1 divides 11.\n        Its double 2^((10*n)+1)-2 exactly divides 11.\n        Hence remainder is 2 in the given question.
913334
x^2 + 2x + 7 Remainder 10\n\ni can add the steps if you don't think i did it by hand
30589
2(5)-5(5)+2(5)=\n10-25+10\n20-25\n-5
737484
5^-2 = (1/5)^2 = (1/5)*(1/5) = (1*1)/(5*5) = 1/25
781604
OK, there's 2 possibilities:\nIf you meant:\nlog (16^1/4), then you need to know the property of logarithms where:\nlog a^b = b log a.\n\nSo:\nlog 16^(1/4) = 1/4 log 16\nlog 10^(.01) = .01 log 10\nlog 2^(1/8) = 1/8 log 2\n\nIf you meant (log 16)^1/4, then I have no idea how to do it by hand.\n\nEdit:\nFurther, you can also do things like:\nlog 16 = log 2^4 = 4 log 2\nso 1/4 log 16 = 1/4 * 4 log 2 = log 2.
803653
2x>-10\ndividing by 2\nx>-10/2\n=x>-5
167287
= (lnx + lnx)-lny = 6+6-2 = 10
462623
>> 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)