Express as a product: log Y to the power of 7, and Y is not a base.?
966617
LogY^7\n\n=7LogY  due to laws of logrithm that LogX^n to any base is equal to nLogX to the same base\n\nThus is a product as 7*LogY\n\nY is not the base. The base is usually writen as a subscript after the Y unless if it is 10. As there is no base writen, the base is 10. Calculators automatically use base 10.
7970
Log(a^b)= b*Log(a)\nLog(ab) = Log(a)*Log(b)\n\na) Log147= ?\n147= 3*7*7\n=> Log147= Log3 * (Log7)^2\n=> Log147= 2yw\n\nb) Log28= ?\n28= 2*2*7\n=> Log28= (Log2)^2 * Log7\n=> Log28= 2xw
812767
The logarithm is the mathematical operation that is the inverse of exponentiation, so you were not wrong.\n\nFor  y = log(base b)(x)\nThe inverse of a logarithm is called an antilogarithm or antilog; thus, x is the antilog of y in the base b. \ni.e x = b^y\n\nSimillarly in a prticular case of natural logarithm (base e)\n\nFor  y = log(base e)(x)\nx is the antilog of y in the base e. \ni.e x = e^y.\n\nI hope, I answere your question.
333624
(ln x)^7=(7*(lnx)^6)*(1/x)=7*(lnx)^6/x
246299
h= (C*I/A+T/k)*z*log(.7*y)
445985
if by this you mean\n\n((3x^-2 * y)^-2)/((4xy^-2)^-1)\n\nthis becomes\n\n((3^-2) * x^(-2 * -2) * y^-2)/((4^-1) * x^-1 * y^(-2 * -1))\n\n((1/9)x^4 * y^-2)/((1/4) * x^-1 * y^2)\n\n((1/9)/(1/4)) * x^(4 - (-1)) * y^(-2 - 2)\n\n((1/9)*(4/1)) * x^(4 + 1) * y^(-2 + (-2))\n\n(4/9)x^5y^-4\n\nwithout using negative exponents, the answer would be\n\n(4x^5)/(9y^4)
822813
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.
252397
The short answer is, you don't.\n\nThe log is the exponent you put on the base to get the number you want the log of.  base 10 log of 100 is 2, because 10^2=100.  But all powers of 1 are 1, and all powers of 0 are 0.
795110
use the Power function\n\n=power(100,1/7)\n(proof - the value you receive "1.93069....."^7 = 100)\n=power(x,1/y)\n\nx=the number you want to find the y root of\ny=the root power
245399
log base (1/5) of 78125 = -7\n\nlog|(1/5) 78125=-7
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...
157812
[x^2/4][2(log(x) - 1) log(x) + 1]
569087
see http://mathworld.wolfram.com/NaturalLogarithm.html
956834
For log(x+8) to be log x +log 8 ,x+8 must be equal to 8x\nWe can write it algebrically as log(x+8)=logx+log8=log(8x)\n=>8x=x+8\n=>7x=8    or,x=8/7
932114
So you have the formula, and you have the number to plug into it. Are you confused about how to deal with the exponents? \n\nIntensity = 5.4 x 10-10, so\n\nd= 10 * log( 5.4 x 10-10/ 10-12)\nd= 10 * log(5.4 / 10-2 )  \nd= 10 * log(540)\n\nd= 10 * 2.732                        note it is a base 10 log!!\nd= 27.32
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.
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.