Logarithm with base e is called_______.?
338108
A Natural log. It is expressed as: ln()\nFor example: ln(e^2) = 2
750081
log (ab)= log a + log b\nlog (a^p) = p log (a)\nlog (a/b) = log a - log b !
815187
5^4=625\nsubscript is the base number, the answer is the exponent, and the regular number after the log (sorry can't remember the name) is the answer.
985432
well...ln(x) does not cancel e^x....but...\n\ngiven e^x, if we take the natural log of e^x then we have ln(e^x), well this means that we have an exponent...we recall that we can pull the exponent down infront of a logrithm...so we now have xln(e)\n\nah! the ln(e) is 1, we recall that ln's are base e and if we take the logrithm of the base we get 1. so we have x*1=x\n\nso...likewise...lnx^2 is 2ln(x)...and thats all we can do\n\nhope this helps\n\nmatttlocke
978189
e^(ln5)=x   .  Now express this in log form by thinking of ln(5)\n                       as just an exponent.\n\n   lnx=ln5   so obviously\n\n   x=5.
855011
log[4]a = k, where [4] is the base.\n\nTherefore, a = 4^k\n\nNow take the log of both sides to the base [a].\n\nlog[a]a = k * log[a]4\n\n1 = k * log[a]4\n\nlog[a]4 = 1 / k\n\nThis is not quite the same as you have, so have you asked to solve the correct expression?\n\nIf, from a = 4^k, you take the log to base [k], it becomes :\n\nlog[k]a = k * log[k]4\n\nSo, log[k]4 = log[k]a / k, instead of your answer 1/ k.
148470
To quote one of my old teachers, who made a lasting impression by screaming the following at our class:\n\nLOGRITHMS ARE EXPONENTS!!!!!!\n\nIf you have an exponent, then you are taking some number to a power.  That number is known as the base.\n\nYou can use any positive number for the base but usually log refers to log base 10. ln is always base e, where e = 2.7...\n\nThat is, saying ln(x) = y is the same thing as saying x = e^y. (y is a logrithm so y is an exponent!)\n\nHistory:\nNatural logs originate in the search for an integral for 1/x. \nIf you define a function, F(x), such that F'(x) = 1/x and start looking at its properties, you find that F(ab) = F(a) + F(b) and F(a^n) = nF(a). The behavior you would expect from a logrithm function. When you calculate the base for that function, you get e.
627488
you can change the base...\nln x = (log2 (x))(ln 2)
689473
get the n's on one side\nso\n17959/21558=(1.02/1.03)^n\nor\n.833055014=.990291262^n\nusing logs with change of base\nn=log (.833055014)/log(.990291262)\nor n=18.72
734891
You can also use Google as a quick calculator.\nJust type in, say, log(6.54321)\nand the answer pops up as 0.815790859.
140923
The other property of logarithms is that Log_x A - Log_x B = Log_x (A/B)\n\n\nso you can combine this rule with the other one i told you about in your other question...\n\nthe only thing is that u are working in reverse order\n\n\nLn 2y/x= Ln 2+ Ln y - Ln x\n\n\nThese are properties, they should be tought to you as principals, without them you cant solve logs:D hope this helps ;) if you think it does pls gimme 10 pts:P
437664
The secret is to understand some of the rules of logarithms and adjust translate the numbers to make your lob easier.\n\nThe rules:\nlog(a*b) = log(a) + log(b)\nlog(a/b) = log(a)- log(b)\nlog(a^n)=nlog(a)\n\nIf I understand your problem:\n\nx= [(86400)*sqrt(365.25)]/[ (.079)^11]\n\nso log(x) = log [.....]\n\n     = log(86400)+log(356.25^.5) - log(0.79^11)\n     = log(86400)+.5*log(356.25) - 11*log(0.79)\n\nuse log tables to get the logs and add them up to get the log of x.  Use the tables again to find x.\n\nYou can make other approximations, i.e., 356.25 is close to 19^2 (=361)\nand 86400 is close to 300^2 (=90000) and continue from there.  \n\nHope that helps.
708168
alog(x) = log(x^a)\n\n1/4log_b(16) = log_b(16^1/4) = log_b(2)\n\n2log_b(5) = log_b(5^2) = log_b(25)\n\nlog_b(7) doesn't change.\n\nlog(x) + log(y) = log(xy)\n\nand \n\nlog(x) - log(y) = log(x/y)\n\nTherefore\n\nlog_b(2)-log_b(25)+log_b(7) = log_b(14/25)
894472
Use a calculator.  Enter [2] [log] for the first one.
1132233
this question will require calcs using logarithm.\nhence will take time
106841
Try the following resources:\n\n1. http://www.krysstal.com/logarithms.html\n2. http://www.physics.uoguelph.ca/tutorials/LOG/\n1. http://en.wikipedia.org/wiki/Logarithms
978128
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.