Anybody know how to do the inverse of log(logarithm)?
548412
If the base is 10, then\n\nantilog x = 10^x\n\nIf the base is e\n\nantilog x = e^x\n\nin general\n\nantilog x = base^x
627488
you can change the base...\nln x = (log2 (x))(ln 2)
941613
log_a b = x is the number to which you need to raise a to get b:\na^x =b\n...........
704178
I'm sure you must have been given these few principles of log operations:\n1. log a + log b = log ab\n2. log a  - log b = log (a/b)\n3. log a^p          = p*log a\n4. log 1 = 0\n5. log (subscript b) b = 1\n\nIn the first problem, if you apply rule 3 it becomes\n1/5(log(x+1)^3 + log(x-1)^2 - log 7), then apply rules 1 and 2 and you get\n1/5log(((x+1^3)*(x-1)^2)/7), a horrible looking expression and I don't know why anyone would ask you to touch it!\n\nIn the second one, log (subscript b) x*b\n= log x + log b [by rule 1]\n= log x + 1     [rule 5]\nTo deal with log sqrt 10, remember \nsqrt 10 = 10^(1/2) and use rule 3;\nalso 10 = 2*5 and use rule 1 in reverse, i.e. split log 10 into log 2 and log 5.\n\nI can't do much more without actually doing them all for you, and I don't think I should do that.\n\nIn  problem 3, take the log of each side and use rule 3, i.e.\nto solve  a^(x + b) = c, do\nlog a^(x + b) = log c\nand then (x + b)*log a = log c\n\nYou can';t do q.4 without knowing what the base of logs is.  I'd assume it's 10 and use rule 1 and solve\nlog x(x+3) = log 10\n\nIn problem 7, I think those 4's are all subscripts, in that case you are right and have used the rule I omitted: \n\nlog (subscript b) x  = (log x)/(log b)\n\nHowever, if you apply rule 3 to log 64, using 64 = 4^3, you'll simplify your answer a great deal.  Is it 3?
338108
A Natural log. It is expressed as: ln()\nFor example: ln(e^2) = 2
921542
Logarithms are exponents, so they follow the rules of exponents.  For instance,\n\n(x^a)*(x^b) = x^(a + b)\n\nTherefore, \n\nlog a + log b = log(ab)\n\nAlso, every exponential equation has a logarithmic form, and vice versa.  So, assuming we're talking base 10,\n\nlog x = y\n\nmeans the same as\n\n10^y = x\n\nSo,\n\nlog a + log b = log (ab)\n\nIn your case:\n\nlog(x) + log(x + 15) = 2\n\nlog(x(x+15)) = 2\n\nIn exponential form, this becomes\n\nx(x + 15) = 10^2\n\nx^2 + 15x = 100\n\nx^2 + 15x - 100 = 0\n\n(x + 20)(x - 5) = 0\n\nx = -20 or x = 5\n\nSince you can't take the log of a negative number, we reject x = -20, so the answer is\n\nx = 5.\n\nDoes that help?
827167
log 2 = 0.3010(base10)
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.
750081
log (ab)= log a + log b\nlog (a^p) = p log (a)\nlog (a/b) = log a - log b !
1140302
Here is an interactive page that discusses one way to do it:\n\nhttp://dilc.upd.edu.ph/loDeploy/math/math10a.html\n\nFor large numbers, bequalmings technique using logs is an easier way to go.
760330
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}
1075997
Since you use log rather than ln, I assuming base 10 (common) logs.\n\nLogrithms are exponents!!!!!\nThey are the exponent you hang on the base to get the number in question.  That is,  saying   log x = y is the same as saying 10^y = x.\n\nSo log x = -2 is the same as saying 10^(-2) = x  or x = 1/100.
914213
OK. It looks like you took the log base 10 of 9 (which is .9542) and multiplied it by 1/81 (which is .01234) to get your .01178. Thus, you found\n(1/81)*Log(9).\n\nTo get log base 9 of 1/81, i.e.\nlog_9 (1/81)\nyou have to find the exponent x where\n9^x=1/81.\nThis exponent will be -2 since\n9^(-2)=1/9^2=1/81.\nNo calculator is needed here!\n\nTo find the result on a calculator, you have to find\nLog(1/81) (which is -1.908485)\nand divide it by\nLog(9) (which is .9542425)\nto get -2.
658763
Take the logarithm of both sides:\n\nlog a^a = log 2\n\nThen use the rule log (x^y) = y log x\n\nso a log a = log 2\n\nUnfortunately from here we can only solve numerically. Look at the graph of a log a and make approximations.\n\nThough not expressible using elementary functions, the actual answer is defined by the Lambert's W function and reads: log 2 / (LambertW (log 2)), which is roughly 1.5596. More about the W function and what it's used for below.
807860
Yes, though don't forget the y on the second log:\n\nlogb x^2/3 + logb y^3/4\n\nAlso, if you would have had:\nlogb x^2/3 / y^3/4 it would equal logb x^2/3 - logb y^3/4\n\nlog rules are just like exponent rules: multiplication = addition and division = subtraction
790275
log(9)3\n=\nlog(9)9^1/2\n=\n1/2*log(9)9\n=\n1/2\n[as log(m)m=1]
659018
Yes, there is a formula to change the base of a logarithm, and here it is...\n\nIf you have an expression log(x) with base 3, the change of base formula allows you to re-express this in the form (lg x)/(lg 3), where lg refers to a logarithm with a base of 10.\n\nFor information on the change of base formula and other logarithmic rules, you may find this page useful:--\nhttp://en.wikipedia.org/wiki/Logarithmic_identities