write the expression log a(y+5)+2log a(x+1) as one logarithm?
815091
First of all, we apply the power rule:  a*log(x) = log(x)^a\n\n     2*log(a(x+1)) = log(a(x+1))^2\n\nNext, we know that the sum of two log expressions can be written as a single product: log(a) + log(b) = log(a*b)\n\n     log(a(y+5)) + log(a(x+1))^2 = log[a(y+5)*(a(x+1))^2]\n\nYou may try some algebraic simplification, but I think that these concepts are the ones that are being tested for this problem.
943827
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...
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
808711
Welcome!!\n\nHere is the solution:\n\nlog(x+1) = log3 - log(2x-1)\nlog(x+1) + log(2x-1) = log3\nlog[(x+1)(2x-1)] = log 3\n\ntherefore,\n\n(x+1)(2x-1) = 3\n2x^2 - x + 2x -1 = 3\n2x^2 + x - 4 = 0\n\nyou need to use the quadratic formula to solve this:\n\nx = {-1 ± sqrt[1 - 4(2)(-4)] } / 2*2\nx =[ -1 ± sqrt(33)] / 4
982509
Let log2 5 = x\n\nThen 2^x must be 5\n\nTake log of both sides but use log10 (common log, on calculator)\n\nx log 2 = log 5\n\nDivide both sides by log 2
1173749
2\n\nwhen adding logs with similar base you multiply; 4*9 = 36\nlog (6) 36 = 2
574952
Use your logorithm rules to combine the terms on the lefthand side into one term\n\nlog (x^2/(8x)) = log (x-10)\n\nNow, if the logorithms are equal, the arguments must be equal\n\nx^2/(8x) = x-10\n\nCross-multiply\n\nx^2 = 8x^2-80x\n\n7x^2-80x=0\n\nx(7x-80)=0\n\nSolving this, you get x=0 or x=80/7. but you have to reject x=0 because it's not in the domain of the original expression (you can't have log(0)).\n\nso x=80/7
871412
Because 10^1 = 10, you know that log(10) = 1.\n\nYou can simplify log(5) / log(10) to:\n= log(5) / 1\n= log(5).\n\nThe final part I can't do without a calculator... it's 0.69897.
850657
In the logarithm, there are several number of rules you have to remember.  One of them is by combining two or more logarithms into one.\n\nFor example,\nlog (a) b + log (a) c = log (a) bc\nlog (a) b - log (a) c = log (a) (b/c)\na is the base of the logarithm.\nb and c is any value more than 0.\nThis kind of combination can only be done if and only if the logarithms have the same base.\n\nThe other rule involved in solving your question is as following:\nc log (a) b = log (a) b^c\na is the base of the logarithm.\nb is any value more than 0.\nc is any real number.\n\nIn your case,\nlog (a) (y + 5) + 2 log (a) (x + 1)\n= log (a) (y + 5) + log (a) (x + 1)^2\n= log (a) (y + 5)(x + 1)^2\n\nIf you would like to multiply it, \nlog (a) (y + 5)(x + 1)^2\n= log (a) (y + 5)(x^2 + 2x + 1)\n= log (a) (x^2y + 2xy + y + 5x^2 + 10x +5)\n= log (a) (x^2y + 5x^2 + 2xy + y + 10x +5)\n\nHowever, this only further complicate the answer, therefore log (a) (y + 5)(x + 1)^2 would be the best answer.
1031270
x^2 + 5 + 5x + x y +5y\n\nx^2 + xy + 5x + 5y + 5\n\nx(x+y) + 5(x+y) + 5\n\nx(x+y) + 5(x+y+1)       OR     (x+5)(x+y)+5\n\nCynthia
725458
Hello dear Sherri \n\n● 4^(2x+1) = 2^(3x+6)\n-Take ' log ' from each sides\n\n● log [4 ^ (2x+1)] = log [2 ^ (3x+6) ]\n● (2x+1) log 4 = (3x+6) log 2\n   { as you know you can get log 4 =  2 log 2 }...\n● 2(2x+1) (log 2) = 3x+6 (log 2)\n   { cancel " log 2 " } ...\n● 4x+2 = 3x+6\n- Now its a simple function \n● 4x - 3x = 6 - 2\n● x = +4\n\nGood Luck Darling ♣
824419
log 10 * log 10 = 1 because log (base 10) 10 = 1 \nand therefore 1 * 1 = 1\n\nlog 10 + log 10 = 1 + 1 = 2 \nand log 10 + log 10 = log (10*10) = log 100\nlog 100 = 2 means 10^2 = 100 noting again that log means log (base 10) meaning it is the power to which the base is raised to get its argument\n\n1000 = 10^3 so log 1000 = 3\n\nNow using the rules stated\n[ie  log a - log b = log (a/b)\nlog a + log b = log ab\nlog a^b = bloga]\n\nlog 3 + 3 log y = log3 + log (y^5) (since bloga = log a^b)\n= log(3 * y^5) (since log a + log b = log ab)\n= log (3y^5)\n\nThe log of a product = the SUM of the individual logarithms!\n\nlog(ab) = loga PLUS logb defintely NOT loga TIMES log b
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
1027143
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)^
960996
log y = d log (cx)\n\nlogy = d log c + d logx\n\nlet logx = X\nlog y = Y\n\nand d logc = C\n\nthe equation becomes\n\nY=dX+C\n\nthis is of the form y=mx+c which is a straight line
1164623
s = c(1 - d)^n\n(1 - d)^n = (s/c)\nlog(1 - d)(s/c) = n\n\nANS : n = log(1 - d)(s/c) or n = log(1 - d)s - log(1 - d)c\n\nWhereas (1 - d) is the base\n\nor you can type it like this\n\nn = (log(s/c))/(log(1 - d))\nor \nn = (log(s) - log(c))/(log(1 - d))
362293
The logarithm (see attached article 1) of a number x in base b is the number n such that b^n = x. It is usually written as log_b(x) = n (b in a subscript).\n\nFor example, log (100) = log_10 (100) = 2, because 10^2 = 100.\n\nThere is one special case - natural logarithm (in base e =~ 2.7182818). It has some properties, which are very useful in calculus. They are written as ln (x).\n\n\nAntilogarithm is a synonym of exponentiation or power.\nThis is an inverse function to logarithm.\nIf log_b(x) = y, then b^y=x.\nFor integer x, b^x mean multiplication b*b*b*...*b (x times).\nFor all real x, b^x are defined in the second attached article, part "Exponentiation with real exponents".\n\nHope it helps.