How to solve 10 = -20[log(x)]?
968865
10 = -20 log x\nlog x = 10/-20 = -2\n10^(log x) = 10^(-2)\nx = 10^(-2) = 1/10^2 = 1/100
829854
Use logarithms to bring exponents down where you can deal with them.  Since the natural log is base e, that's the way to go here.\n\ne^5x = 7\n\n5x = ln 7\n\nand solve from there.
382118
(10^2-10^2)/10(10-10)\n= (10+10)10-10)/10(10-10)\n= 20/10 = 2
634484
log_2(X) + log_2(X+5) = 1\nlog_2(X * (X+5)) = 1\n2 ^ log_2(X * (X+5)) = 2^1\nX * (X+5) = 2\nX^2 + 5*X = 2\nX^2 + 5*X - 2 = 0\nX = (-5 +- sqrt(5^2 + 4*1*2)) / 2\nX = (-5 +- sqrt(33)) / 2\n\nSo, two possible answers are:\nX1 = (-5 + sqrt(33)) / 2\nX2 = (-5 - sqrt(33)) / 2\n\nAlso, logarithms have sense only if these inequality are true:\nX > 0\nX + 5 > 0\n\nX2 < 0, then X1 is the only possible answer.\n\nFinally:\nX = (sqrt(33) - 5) / 2
674167
Solve this by using logarithms\nlog((1/5)^2x) = log(625)\n2x*log(1/5) = log(625)  Product Rule\n2x = log(625)/log(1/5)\n2x = -4\nx = -2\n\ntosmith
627488
you can change the base...\nln x = (log2 (x))(ln 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
324025
log 10000 = x\n10^x = 10000\nx = 4\n\n4
555213
x^2 - 20 = 0\nx^2 = 20\nx = ±sqrt(20)\nx = ±sqrt(4 * 5)\nx = ±2sqrt(5)
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.
470800
logx 1/a = (logx 1) - (logx a)\n= 0 - 3\n= -3
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
399448
log(3x+7) = 0\n10^0 = 3x+7\n1 = 3x+7\n-6 = 3x\nx = -2
427245
You can only take the log of numbers that are positive, so\n\nx^2 + 6x + 5 > 0\n\nSo (x+5)(x+1) > 0\n\nSo x would have to be greater than -1 and less than -5.
1046114
80=16*5\n\nlog(2^4*5)/log2\n=(log2^4+log5)/log2\n=(4 log2+log(10/2))/log2\n=(4log2+log10-log2)/log2\n=(3log2+1)/log2=3+(1/log2)
593752
log ( sqrt (x+ 1) *x^5)
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