I am sorry but i cant write it out like it really is in the book. so in words its log2 parenthesis 6x-4 over x+1 parenthesis..then equals 4 ? Sherman if you are out there you are GREAT !! HELP
133180
METHOD 1:\n\nlog2[(6x-4)/(x+1)] = 4\n\nConvert to Exponential form:\n\n2^4 = (6x-4)/(x+1)\n\n16 = (6x-4)/(x+1)\n\nCross Multiply:\n\n16(x + 1) = 6x - 4\n16x + 16 = 6x - 4\n16x - 6x = -4 - 16\n10x = -20\n\nx = -2 ANSWER\n\nCheck:\n\nlog2[(6x-4)/(x+1)] = 4\nlog2[((6)(-2)-4)/(-2+1) ?=? 4\nlog2[(-12-4)/(-1) ?=? 4\nlog2[(-16)/(-1) ?=? 4\nlog2(16) ?=? 4\n4 = 4\n\nMETHOD 2:\n\nLog2[(6x-4)/(x+1)] = 4\n\nConvert to Base 10 Log. (Common Log.):\n\n(log[(6x-4)/(x+1)])/(log2) = 4\n\nCross Multiply:\n\nlog[(6x-4)/(x+1)] = 4(log2)\nlog[(6x-4)/(x+1)] = log(2^4)\nlog[(6x-4)/(x+1)] = log 16\n\nRemove the Logarithms:\n\n(6x-4)/(x+1) = 16\n\n6x - 4 = 16(x + 1)\n6x - 4 = 16x + 16\n6x - 16x = 16 + 4\n\n-10x = 20\n\nx = -2 ANSWER
459505
In the first problem, the 15/x - 15/x cancels out, so the equation is true whenever x is not 0.  x>0, x<0.\n\nIf there's a parenthesis around the (x+2), that's different.\n15/x - 15/(x+2) = 2\nMultiply everything by x(x+2)\n15x(x+2)/x - 15x(x+2)/(x=2) = 2x(x+2)\n15(x+2) - 15x = 2x(x+2)\n15x +30 -15x = 2x^2 +2x\n0 = 2x^2 +2x -30\n0 = x^2 +x -15\nYou should be able to get it from there.\n\nSecond one, assuming the parentheses are missing?\n3x/(x+2) - x/(x-2) +8/(x^2 -4) = something (zero?)\nMultiply everything by (x^2-4)\n3x(x^2-4)/(x+2) - x(x^2-4)/(x-2) + 8(x^2-4)/(x^2-4) = 0\nRecall x^2 -4 = (x+2)(x-2)\nSo factoring\n3x(x+2)(x-2)/(x+2) - x(x+2)(x-2)/(x-2) +8 = 0\n3x(x-2) - x(x+2) +8 = 0\n3x^2 -2x -x^2 -2x +8 = 0\n2x^2 -4x +8 = 0\nx^2 - 2x +4 = 0\nShould be able to get home from there.
230456
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
315513
6x - 4x -3x = 5x becomes 6x - (4x -3x) = 5x\n6x - 4x -3x=`x I'm not sure about this one.  Is it supposed to be 6x - 4x -3x = -x? then it would be (6x-4x) - 3x = -x\n7 + 5p - p = 11p becomes (7 + 5)p - p = 11p\n7 + 5p - p = 7 becomes 7 + 5(p - p) = 7\n7 + 5p - p = 7 + 4p becomes 7 + (5p - p) = 7 + 4p\nYou do not actually need parentheses for the previous one.\n7 + 5p - p = 0 I cannot figure this one out.  Is something typed wrong?
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:)
55023
i think i know how to do it, one minute...\nhey is it 7k-14/(2k-6)(k-4) -6K+18/(2K-6)(K-4) = K+4/(2K-6)(K-4)\nor (7k-14)/(2k-6)(k-4) -(6K+18)/(2K-6)(K-4) = (K+4)/(2K-6)(K-4)\ni added parenthesses\n\n>>>ok so that's what you need to do:\nlet's say (2k-6)(k+4) = X\nso you get:\n7k -14/X -6k +18/X = k +4/X\n7k -6k -k = 4/X +14/X +18/X\n0k = 0X\n0 = 0*(2k-6)(k-4)\n=> every k will give us a legal equation!\n\ni really hope it's true, have u got the answer?
422410
Do you know your log identities?  That would help you here a lot!\n\nUse these log identities to solve your problems:\n\n. log(a) + log(b) = log(ab)\n. log(a) - log(b) = log(a/b)\n. c*log(a) = log(a^c)\n\nSo, for problem 1 (I'm assuming the "a" means that "a" is your base) we get the following:\n\nlog a(y+5)+2log a (x+1) =\nloga(y+5) + loga((x+1)²) =\nloga((y+5)(x+1)²)  \n\nProblem 2:\n\nIn(x+5)-In(3)=In(x-3)\nln((x+5)/3) = ln(x-3)\n(x+5) / 3 = x-3\nx + 5 = 3x - 9\n2x = 14\nx = 7\n\nProblem 3:\n\n3^x+5=9^x\n\nIs 3 raised to the power of x only, or to the power of x+5?  It's hard to tell, and it does matter to the solution.\n\n~ ♥ ~
523558
He should have put parenthesis around the 3X5^Y and then he was expanding the log(3X5^Y) to the log3 + log 5^Y because the sum of the logs is equal to the log of the products.  The log 5^Y can be written as Y(log 5).  You end up with\nY(log2) + log2     = log3 + Y(log5)\nY(log2) - Y(log5) = log3 - log2\nY(log2 - log5)      = log3 - log2\nY= (log3 - log2) / (log2 -log5)\nY= (log (3/2) / log(2/5) = log(1.5) / log(.4)\nY approximately = -.4425\nyou can check by substituting back in the original equation\n2^(-.4425 + 1) = 3 * 5^(-.4425) (use the reciprocal of 2nd factor\n2^.5575           = 3 / 5^(.4425)    because of the negative exponent)\n1.4717...         = 1.4717...
46795
...=(logA+logA)-(logB-logB-logB)+1/5logC=(logA^2-logB^3)+1/5logC\n= log(A^3/B^3)+1/5logC =log(A^3/B^3)+logC^1/5=log[(A^3*C^1/5)/B^3]\nNow if you know to transform C^1/5 (i don't really remember now)it becomes even simpler.
494377
log 6^(2x)= log 9037984\n\n2x*log 6= log 9037984\n\n2x= log 9037984/log 6\n\nx= log 9037984/(2*log 6)\n\nx= log 9037984/log 6²\n\nx= log 9037984 / log 36\n\n₢
462199
Stop, stop, stop. If it was f(x)=2logx, f(x) cannot be = to (1/log2)*logx, as  1/log2 is not = 2.\nDon’t hurry, put your question correctly with necessary details and don’t save on ()-s in formulas.
433445
Well the log is the inverse of an exponent right?\n\nLog base B (X) = N is equivilant to \nB^N = X\n\nYou want to find why Log 2 = .301 right?\nLet's turn it into an exponent.  I assume you're using log base 10 so we get:\n10^.301 = 2 (approximately) right?\nThis is because .301 is the exponent that you must bring 10 to the power of in order to get 2.  That's the best explanation I can offer.
387012
First let me assert that \n  log(base b)(a)=1/log(base a)(b) as follows:\n\n let log(base a)b=z\n   a^z=b\n   a=b^(1/z)  . Now sub back for z giving\n   a=b^(1/log(base a)(b))  and convert to log form of base b\n 1)   log(base b)(a)=1/log(base a)(b)  \n\n Now follow this: \n    let  log(base c)(x)=y\n        c^y=x   Now take log(base a) of both sides\n    ylog(base a)c=log(base a)x  and  sub back for y\n log(base a)x  = log(base c)x{log(base a)c} and\n use 1) to convert log(base a)c giving\n   2)  log(base a)x  = log(base c)x/{log(base c)a.\n\n     Lastly, into 2)  substitute X^3+1 for x,  4 for a and 10 for c\n\nand we have \n  log(base 4)(x^3+1)  = log(base 10)(x^3+1)/{log(base 10)4}
522151
log(3x) - 3[log(x) - log(1+y)]\n= log(3x) - 3log[x/(1+y)]\n= log(3x) - log[x^3/(1+y)^3]\n= log[ (3x) / (x^3/(1+y)^3)]\n= log[ (3x) * (1+y)^3 / x^3)\n= log(3(1+y)^3 / x^2)
217740
(1 + 9) x (5 x 2) + (6 + 8) - (3 + 4 + 7) \n\nsome of those parentheses are superfluous, but they make it easier to see how it works.
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The order of operations is PEMDAS, best remembered by "Please Excuse My Dear Aunt Sally".  It stands for Parentheses, Exponents, Multiply, Divide, Add, Subtract.\n\nUsing this, you would solve the equation as follows:\n\n1) Parentheses: Take (x2-3y) first. Insert x and y values to get (-1 x 2 - 3 x -2).  Based on the order above, you need to multiply within the parentheses first. So you take -1 x 2 = -2.  3 x -2 = -6.  This leaves you with -2 - -6.  Subtracting a negative gives you a positive, so you have -2 + 6= 4.  The answer to the parentheses is 4.\n\n2) Multiply the end of the equation, the -4y3.  This becomes -4 x -2 x 3.  This gives 24.  (negative times negative equals positive).\n\n3) Before you divide, simplify the right side by combining like terms. 5 + x + 24.  So, 5 + -1 + 24 = 28.\n\n4) You now have 4/28.  This reduces to 1/7. That is the answer, 1/7.  (Sorry I had it wrong before.)
362110
you gotta get it to two sets of parentheses, i don't remember exactly how to do it, but i know it should look like (x+1)(x+1) that's not the right answer, but i kinda remember doing this in school