How is logarithmic calculations used in a civil engineering career?
441387
There was a time when all good civil engineers had to know how to operate a slide rule (which used sliding logarithmic scales to add logs (multiply) and subtract logs (divide), etc.  Civil engineers must be familiar with the fundamentals of all engineering disciplines and would know that the pH of acids and alkalies are logarithmic, the dB of sound decibels are logarithmic and the rickter scale used to measure earthquakes (and to design structures to withstand them) is logarithmic.  They must know the significance (and use) of logs to base 10 and natural logs to base e to work with advanced formulas and calculations.
47515
There are lots of ways to find logarithms without a calculator. After all, people had to do this when there were no calculators. \n\n1. Logarithm tables are in books and all over the place. Here is an example of a log table on the internet:\n\nhttp://www.sosmath.com/tables/logtable/logtable.html\n\n2. My advisor in graduate school actually memorized a lot of values from the log tables so he could do very fast mental calculations. \n\n3. Slide rules are easy ways to get approximate logarithm values. \n\n4. There are numerical methods for approximating logarithms that you can track down, for various values of the argument. For example, look at \n\nhttp://www.maths.abdn.ac.uk/~igc/tch/ma1002/appl/node59.html\n\nhttp://mathworld.wolfram.com/NaturalLogarithm.html\n\nhttp://www.informatik.tu-darmstadt.de/ftp/pub/TI/reports/MarkusMaurer.diss.pdf\n\nAfter all, the calculator has to use some method to produce or approximate the logarithm, and all one has to do is to use similar methods to find the logarithm by hand, as long as they do not require an excessive amount of computing.
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.
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.
205936
1) The natural logarithm started life as the integral of 1/t from 1 to x. This integral was defined to be the natural logarithm because it satisfied the properties of a logarithm in x. The base of this logarithm was called e, Euler's number. Describing the usefulness of this number, and thus why everyone accepted the term natural logarithm, fills several good books, but you can find a condensed version in the source link. \n2) A distribution which is log-normal may be written as a large sum over the logarithm of some random variable, and thus the logarithm of a product over that random variable. Consider the central limit theorem applied to the logarithm of a random variable. More information can be found in the source link below.
713982
use logarthmic differentiation\n\ny = x^(1/x)\n\nln(y) = (1/x)* ln(x)\n\ndiff w.r.t x\n\n(1/y) (dy/dx) = (1/x)*(1/x) +ln(x)*(-1/x^2)\n\n(dy/dx) = { y/(x^2) }  [ 1 - ln(x) ]\n\nsubstitute for y
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.
496208
log(x) - log(1 + (x^2 + 1)^(1/2))\n\nWhere log is the natural logarithm, of course.
757898
Remember that with logarithms, exponents within the log become the coefficient outside the log.\n\n1.) logc (sqrt x)\n= logc (x^1/2)\n= (1/2) logc x\n= (1/2) * 1.2\n= 0.6\n\nIf numbers are being multiplied within the log, they can be split up to the sum of the log of the numbers.  It's a similar rule to that in problem 1, as you can imagine:\n\nlog (x^2) = log (x * x) = log x + log x = 2 log x\n\n2.) logc xy\n= logc x +logc y\n= 1.2 + 3.9\n= 5.1
968058
Type 1.16 log / 4 / 1.0125 log =\nThis is 3 approximately (2.9869...)\n\nTh.
881778
(1.0442) to the power n(number of years)=1.7, the ratio of the final to the original amounts. \nFrom logarithms, n log (1.0442) = log 1.7. considering the same base (let us say 10).\nlog 1.7= 0.2304489214\nlog 1.0442= 0.017867719\nhence, log1.7/log1.0442= 12.2685657176, say approximately in 12 years.
962408
log4+xlogb=xloga\nx(loga-logb)=log4\nx=log4/(loga-logb)
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.
90904
A slide rule is an adding which consists of two strips of wood with a movable center strip.  Various functions are placed on the strips according to their logarithms/  It works according to the idea that multiplication is done by adding logarithms and division is done by subtraction of logarithms.  Accuracy is to 3 significant digits.  It does have the advantage that it doesn't need a battery to work!
519146
Engineering uses a lot of mathematical procedures - and with some physics background too i would have thought with a short training period you would have been an ideal candidate.\n\nI am a maths graduate and my husband an engineering graduate and there are always takers for people with those kinds of qualifications. (beats BA 'leisure studies' dont you think?)
1128770
You want to exponentiate your logarathim. For example, a=log x is the equivalent of 10^a=x.\n\nIn your problem, first multiply both sides by negative one to move the negative sign to the other side. Then, -6.00=log[OH-] becomes 10^(-6)=[OH-]. Your concentration of OH- is 10^-6 M.
840849
1.52582219663167... if the log is base e\n1.59874065990736... if the log is base 10\n\nAs the use of a graphing calculator does not give any insight into the derivation of the solution, here it is:\n\nTo solve this, you need to use some numerical interpolation method, as the equation can not be solved directly for x (in terms of finitely many algebraic functions).\n\nNewton's method yields:\nLet y = f(x) = log(x) + 3x - 5 \n{here log(x) is the natural log(base e).  We want to find the value of x for which f(x) = 0}\nf '(x) = 1/x + 3\nthe method says that x(n+1) = x(n) + (y - f(x(n)) / f '(x(n))\nand here, y = 0, so\n\nx(n+1) = x(n) + (5 - log(x(n)) - 3x(n)) / (1/x(n) + 3)\n\nthen, x = lim(n→∞) x(n)\n\n(You must choose something reasonably close for x(1), such as\nx(1) = 1.  It takes only five iterations to get the accuracy of my solutions above in a VB6 program using double precision variables.  i.e., x(5)=1.52582219663167)