The fours are the bases, they are small then the rest\n\nlog4 (x-3)+log4 (x+3)=2
120762
The sum of the logs is the log of the products, ie.\nlog(a)+log(b)=log(ab)\n\nSo combine the sum of logs here into a single log of the product of (x-3) and (x+3).\nlog_4(x-3)+log_4(x+3)\n  =log_4((x-3)(x+3))=2\n\nOnce you have a single log isolated on one side of the euqation you should remember that the log equation log_a(b)=x is equivalent to the exponential equation a^x=b, so in this case\n(x-3)(x+3)=4^2=16\n\nNow you've got a quadratic equation that you can solve for x by expanding everything, moving everything to one side of the equation, and factoring.\nx^2-6x+9=16\nx^2-6x-5=0\n(x-5)(x-1)=0\nx=1 or 5\n\nYou should then check to see if both of these solutions are valid in the original equation. Since letting x=1 would involve taking the log of a negative number in the original equation (which you cannot do if you are dealing with real numbers) you should throw that solution out.\n\nx=5
338546
Logs are just equivalent to power statements. \nlog_a (b) = c means a^c = b.\nSo that equation just means x^2 = 2x-3.\nNow thats just a quadratic;\nx^2 - 2x + 3 = 0\nHowever that has no real roots. So there are no such values of x. But hopefully you get the idea.
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
406589
The inverse of log3 is 3^, so x-2 = 3^4 = 81, so x=83
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)
469334
you can try breaking the problem up into pieces:\n32/4 = 8 (these are just the coefficients)\ny^16/y^4 = y^12 (remember you subtract exponents if the bases are the same and being divided)\nso your final answer is:\n8y^12\n\n\nthere really are no steps! \nhope this helps!
469628
log2 8 = 3, because 2^3 = 8\n\nlog 4 8 = x, then 4^x = 8.  4 is 2^2, so 4^x = 2^2x = 8\n2^2x = 2^3\n2x = 3\nx = 1.5\n\nlog8 8 = 1, since 8^1 = 8\n\nlog16 8 = x\n16^x = 8\n2^4x = 2^3\n4x = 3\nx = 3/4\n\nlog32 8 = x\n32^x = 8\n2^5x = 8\n5x = 8\nx = 8/5\n\nSo we have 3 + 1.5 + 1 + 3/4 + 8/5 = 7.85
549032
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.
79297
x is anything less than 1.\n4 x1=4 so 1 cannot be the answer (unless the < had a line underneath it)\n4x0=0 which is less than 4\n4x -2= -8
503501
(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
549488
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.
62442
[x^2/4][2(log(x) - 1) log(x) + 1]
579884
log 2 x^2+1-log2^x\nlog 2 + log X^ + log 1 - (log 2^ + log X)\n=log 2 + A log X + log 1 - A log 2 - log X\n=log 2 + log 1- log X + A log (X - 2)\n=log (2 x 1) - log x + A log (x -2)\n=log 2/x + A log (x-2)\n=log 2/x + log (x-2)^\n= log (2/x X (x-2)^) = log 2/x(x-2)^
528099
Problem #1:\n2^(3x) = 4\n\nSince you have powers of 2, I would take the log base 2 of both sides:\nlog[2](2^(3x)) = log[2](2²)\n3x = log[2](2²)\n3x = 2\nx = 2/3\n\nProblem #2:\n5/6x = 10\n\nMultiply both sides by x\n5/6 = 10x\n\nDivide both sides by 10:\n5/60 = x\n\nReduce:\nx = 1/12\n\nYour answers are:\n2/3 and 1/12
156734
Let's call f(x) = x^2 - 4\n\nThe graph of f(x) = x^2 - 4 looks like a smile ( U )  with the graph passing the y - axis at -4.\n\nThe roots are at x = 2 and x = -2.  This can be found by solving for x^2 - 4 = 0\nx^2 - 4 = 0\n(x + 2)(x - 2) = 0\nso x + 2 = 0, or x - 2 = 0\nso x = -2 or x = 2\n\nsince you're int'ing from x = -1 to x = 3, you have to break up the int into\n[x = -1 to x = 2, where f(x) is negative and take the absolute value] \nand\n[x = 2 to x = 3, where f(x) is positive]\n\nso int of f(x)\n= x^3/3 - 4x + c\n\nso int of f(x) from x=-1 to x=3\n= {+ve value of int of f(x) from x = -1 to 2}\n                +\n  int of f(x) from x = 2 to 3\n\nSeparately\nint of f(x), x = -1 to 2\n= | (2^3)/3  - 4(2) - [((-1)^3)/3  - 4(-1)] |\n= | (8/3) - 8 - [(-1/3) + 4] |\n= | (8/3) - 8 + (1/3) - 4 |\n= | (9/3) - 12 |\n= | 3 - 12 |\n= | -9 |\n= 9\n\nint f(x), x = 2 to 3\n= (3^3)/3  - 4(3) - [(2^3)/3  - 4(2)]\n= 27/3 - 12 - [(8/3) - 8]\n= 9 - 12 - (8/3) + 8\n= 5 - (8/3)\n= 7/3 (or 2 1/3)\n\nAdd them together and you get:\n9 + 2 1/3\n= 11 1/3 which is appromixately 11.333_
393945
log10^x=4\nTherefore x log10 = 4 (since log(a^n) = nloga)\nlog10 = 1 (assuming log means log (base 10))\nTherefore x = 4\n\nOtherwise x = 4/log10
451886
log3x - 3[logx - log(1+y)]\nwe know \nn log A = log [ A ^n]\nlogA - logB = log [A/B]\nso, going from right to left,\nlog 3x - 3[ log [ x/(1+y)]\nlog3x -  ( log [(x/(1+y)) ^3])\nlog [ 3x / (x/(1+y)) ^3) ]\nlog [ 3 * ( ( 1+y) ^3) / x^2  ]\n:)