a) Determine the domain of log3 (x - 24) ?
439071
a) log3(x-24) is valid for all x but x = 24\nb) log3(x) is valid for all x but x = 0\nc) x-24  = 3^4 solve for x :)\n\nRival
157812
[x^2/4][2(log(x) - 1) log(x) + 1]
627488
you can change the base...\nln x = (log2 (x))(ln 2)
1117027
Log0.78= -0.107905397309519598284728044078063\nLog2=0.301029995663981195213738894724493\nLog0.78/Log2= -0.358453970912476353890087260774269\n\nPut your cursor on the answer to see full length
978128
log(base a) b = logb/loga.\n\nSo take the log of X and divide by the log of 1/3.  Then press the +/- button because you're being asked for the negative of that value.
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.
502102
well chinaren got it started but I believe his explianation is weird.\nLets assume x=3^t\nthen log3(x)+log2(x+1)=3\n=>log3(3^t)+log2(3^t+1)=3\n=>tlog3(3)+log2(3^t+1)=3\n=>log2(3^t+1)=3-t\n=>3^t+1=2^(3-t)\n=>3^t+1=2^3/2^t\n=>(3^t+1)*2^t=2^3\n=>3^t*2^t+2^t=8\n=>6^t+2^t=8\nfrom here there is one one solution possible as t=1\nso x=3^1=3
730130
Domain = x... should be all real numbers\n\nRange = y... 2 and up... x can be a negative number, but will always end up being postive when added to two, therefore the lowest number is 2
980349
Domain:  all real numbers\nRange:  [2, positive infinity]   (all real numbers from 2 to positive infinity)
881529
fist domain\n\nthe number we search the log must be positive so x>-2 \n\ndomain -2<x<infinity \nfor infinity log (base3), infinite is infinite. for x=-2 we tend to x+2 =0 and log (x+2 ) = -infinite  Vertical asypmtote for x=-2 \n\nthe equation can be rewritten  h(x)  = log (base 3) (x+2/7^3)
584413
log(x-4) - log(3x-10) = log(1/x)\nlog((x-4)/(3x-10)) = log(1/x)\n(x-4)/(3x-10) = 1/x\nx(x-4) = 3x-10\nx^2 - 4x - 3x + 10 = 0\nx^2 - 7x + 10 = 0\n(x-2)(x-5) = 0\n\nx = 2 or x = 5.\n\nedit: Sorry, ervin is right.  I thought I had checked, but apparently I wasn't careful enough.  The only solution is x=5 since you cannot take the log of negative numbers.
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)
942101
start by seperating y, u then have y=3x/2+3 then plug in the domain, or the numbers u are givin\nu'll have 0, 1.5, 3, 6, 7.5
689473
get the n's on one side\nso\n17959/21558=(1.02/1.03)^n\nor\n.833055014=.990291262^n\nusing logs with change of base\nn=log (.833055014)/log(.990291262)\nor n=18.72
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
1069254
I think it's log 3x2[6y]/z\n\n*square root of 6y\n*3x squared
694986
domain is the value of x\nthat is {5,-3,4,-6.....}\nrange is the value of y\nthatis{2,4,2,3....}