Is log x + y logx + logy
What is the logarithm function?
|......||The logarithm function is the inverse function of the exponential function with f (x) = ax (a> 0). That means that in the representation y = ax the variables x and y are swapped: x = ay. |
With a fixed base, a power is not assigned to the exponent, but an exponent to the power.
You read loga(x) as Logarithm of x to base a. The logarithm is an exponent.
The equation 9 = 3y leads to y = 2 or log3(9)=2.
The equation 3 = 3y leads to y = 1 or log3(3)=1.
The equation sqrt (3) = 3y leads to y = 1/2 or log3[sqrt (3)] = 1/2.
The equation 1/3 = 3y leads to y = -1 or log3(1/3)= -1.
Three logarithms Top
The logarithm function is important when the base is 2, e (Euler's number), or 10.
According to ISO 31-11, there are simpler spellings for the associated function terms, namely
lb (x) is called binary or dual, ln (x) natural and lg (x) decadic logarithm.
|......||The graphs on the left belong to|
> The logarithmic function has D = | R with the largest possible definition range+ the range of values W = | R.
> The graph is strictly monotonically increasing.
> The y-axis is asymptote.
> The y values are negative for x <1 and positive for x> 1.
> For x towards infinity, y also approaches infinity.
Laws of Calculation Top
The power laws lead to the logarithmic laws (I), (II), (III) and (IV).
(I) log (pq) = log (p) + log (q)
Let h = loga(p) or aH= p and k = loga(q) or ak= q.
Then pq = aH * ak= ah + k .
The power equation ah + k = pq leads to loga(pq) = h + k or loga(pq) = loga(p) + loga(q), wzbw ..
(II) log (p / q) = log (p) -log (q)
Let h = loga(p) or aH= p and k = loga(q) or ak= q.
Then p / q = aH / ak= ah-k .
The power equation ah-k = p / q leads to loga(p / q) = h-k or loga(p / q) = loga(p) -loga(q), wzbw ..
(III) log (xn) = n * log (x)
Let h = n * loga(x) or h / n = loga(x) or ah / n= x.
The power equation ah / n = x leads to aH = xn or lieda(xn) = h or loga(xn) = n * loga(x), wzbw ..
(IV) log (x1 / n) = (1 / n) * log (x)
Let h = (1 / n) * loga(x) or hn = loga(x) or ahn= x.
The power equation ahn = x leads to aH = x1 / n or lieda(x1 / n) = h or loga(x1 / n) = (1 / n) * loga(x), wzbw ..
For example, the question arises as to what relationship exists between the logarithms lg (x) and ln (x).
Claim: lg (x) = ln (x) / ln (10)
It follows from y = lg (x) or y = log10(x) the power equation 10y= x.
Then ln (x) / ln (10) = ln (10y) / ln (10) = y * ln (10) / ln (10) = y. But that means lg (x) = ln (x) / ln (10), wzbw ..
In the same way one can generally show: liedb(x) = loga(x) / loga(b).
Determination of the logarithms Top
In general, the logarithms are transcendent numbers that can be approximated as decimal fractions. These are obtained via convergent series, but then with any precision.
|......||It is not difficult to get logarithms today. There are calculators.|
The TI30, for example, has the LOG and LN buttons for this.
With LN one determines e.g. ln (2) with the key sequence 2 LN and gets ln (2) = 0.693147181.
It makes no sense to take all decimals from the calculator.
Four decimals are probably sufficient, so that the following applies: lg (2) = 0.3010 and ln (2) = 0.6931.
Conversely, if you want to determine the power when the logarithm is given, the key combinations 2nd LOG and 2nd LN are available.
For example, one calculates x = 100,3010 with the key sequence 0.3010 2nd LOG and receives x = 1.99986187.
For example, one calculates x = e0,6931 by the key sequence 0.6931 2nd LN and receives x = 1.999905641.
In both cases, this is rounded to 1.9999 or rounded to make sense 2.
In pre-computer times, the logarithms (for the multiplication of numbers) were read from tables (1).
|......||This is the section of a table. One finds lg (2) = 0.3010. However, the log tables only provide the sequence of digits. You have to set the comma yourself, as shown in the following examples.|
Logarithm as a calculation aid Top
Before calculators existed, decadic logarithms made from tables made it easier to multiply, divide, exponentiate and extract the roots, but at the expense of accuracy.
Adding instead of multiplying
The procedure should be explained using the product 896 * 271.
> From the table you can read off: lg (896) = 2.9523 and lg (271) = 2.4330
> The sum of the logarithms is then 2.9523 + 2.4330 = 5.3853.
> From the table we find backwards 5.3853 = lg (242800).
> Result: 896 * 271 = 242800. The product 896 * 271 = 242816 has been determined with an accuracy of 4 digits.
The equation lg (896 * 271) = lg (896) + lg (271) is the explanation.
If in an equation b = ax Given the power and base, the exponent is calculated by taking the logarithm.
That is, one starts from b = ax to lg (b) = lg (ax) or lg (b) = x * lg (a) and finally to x = lg (b) / log (a) over.
282475249=7x leads to x = lg (282475249) / lg (7) = 8.4510 / 0.8451 = 9.93 (exact value x = 10).
If you determine the logarithms more precisely, e.g. with the TI 30 pocket calculator, you get with
lg (896) + lg (271) = 8.4509804 / 0.84509804 = 10 the exact value.
Multiply with the slide rule
Even before computers existed, the logarithmic tables were replaced by the slide rule. An accuracy of about 3 digits was sufficient.
This multiplication will be explained using the example 121 * 127.
> The 1 of the basic scale of the tongue is placed above 1.21 of the basic scale of the body.
> Set the runner to 1.27 on the scale of the body.
> One reads off the sequence of digits underneath: 1538. (The 8 is uncertain.)
> The approximate calculation of 100 * 100 = 10,000 is used to determine the size of the product.
> Result: 121 * 127 = 15,380. The exact product is 15367.
It is also a matter of adding logarithms. You hang routes one behind the other.
Logarithmic scale Top
If you choose a logarithmic scale for the y-axis, as it appears on the slide rule, the graphs of exponential functions become straight lines.
|......||Taking the logarithm of the functional equation y = 10x, we get lg (y) = lg (2) x or lg (y) = x.|
If one occurs lg (y) against x, one obtains the first bisector as a graph. This is the red line.
The y-axis scale is called the logarithmic scale.
The black straight line is obtained by taking the logarithm of y = ex as lg (y) = lg (e) * x, the green straight line as lg (y) = lg (2) * x.
|......||If a logarithmic scale is used for both axes, the normal parabola results, for example, with p (x) = x2 as a straight line with a slope of 2.|
Take the logarithm of the equation y = x2 , one obtains namely lg (y) = 2 * lg (x).
The advantage of this display is that a wide definition range and thus also a range of values can be displayed.
|......||Because the scales are labeled with powers of 10, a square grid is created.|
With double-logarithmic paper you can see that with a finer division the distances are not the same.
Derivative and antiderivative Top
The following statements refer to the natural logarithm.
|......||The logarithmic function can be differentiated as the inverse function of the exponential function.|
Claim: The derivative of the function with g (x) = ln (x) is g '(x) = 1 / x.
|......||Claim: The antiderivative of the logarithm function is G (x) = x * ln (x) -x + C.|
Interpretation as a patch
|......||The function h (x) = 1 / x is given.|
Then applies to the yellow patch.
The logarithmic function can thus be visualized as the area bounded by the x-axis, the graph of the function h (x) = 1 / x and the verticals x = 1 and x = t.
The graph of the exponential function with f (x) = ex delimits a piece of land in the second quadrant that extends to infinity. Nevertheless, the area is finite, namely 1 FE.
Series developments Top
ln (1 + x) and ln (1-x)
A known series is ln (1 + x) = x- (1/2) x² + (1/3) x³- ... with the convergence range -1
The following geometrical series is used as a starting point.
1 / (1-x) = 1 + x + x2+ x3+ ... + xn-1+ xn/ (1-x)
Then one calculates an integral in two ways.
One can neglect the integral on the far right [(2) page 176f.].
I will not go into this estimate.
So we have ln (1-x) = - x- (1/2) x²- (1/3) x³- .... The series is convergent for -1 <= x <1.
Inserting the boundary value x = 1 in -ln (1-x) = x + (1/2) x² + (1/3) x³ + ... results in the Leibniz series 1+ (1/2) + (1/3) + .... She is divergent.
Replacing x with -x results in ln (1 + x) = x- (1/2) x² + (1/3) x³- ....
The series is convergent for -1
ln (1-x) / ln (1 + x)
Above it says that the logarithms can be calculated using rows. The two series for ln (1-x) and ln (1 + x) are less suitable for the calculation because they only converge slowly
[(3) page 126f. and numerical example below].
But you can combine the two rows.
We have ln [(1 + x) / (1-x)] = 2 [x + (1/3) x3+ (1/5) x5+...].
For -ln (1-x) we write ln [1 / (1-x)].
Then add the series ln [1 / (1-x)] = x + (1/2) x² + (1/3) x³ + ... and ln (1 + x) = x- (1/2) x² + (1/3) x³- ... in terms of terms and receives ln [(1 + x) / (1-x)] = 2x + (2/3) x3+ (2/5) x5+ ... = 2 [x + (1/3) x3+ (1/5) x5+...].
> The sequence to ln (1 + x) gives for x = 2 and 4 summands ln (2) = 1-1 / 2 + 1 / 3-1 / 4 = 0.58.
> The sequence to ln (1-x) / ln (1 + x) gives for x = 1/3 and 3 summands ln (2) = 2 [1/3 + (1/3) (1/3)3+(1/5)(1/3)5] = 0.693004114 or rounded 0.693.
> The table value is ln (2) = 0.693147181.
In this context, the sequence with (1 + 1/2 + 1/3 + ... 1 / n) -ln (n) is noteworthy.
The consequence is convergent with the limit value gamma, the Euler-Mascheroni constant. It has the decimal fraction representation gamma = 0.5772 ....
exp (x), exp (-x), -exp (-x), -exp (x)
-log (x), -log (-x), log (-x), log (x)
Logarithm function on the Internet Top
Logarithms, exercises, results
K. Eckhardt (University of Hohenheim)
Logarithmic scales (.pdf file)
Logarithm, decadic logarithm, exponential function, integral logarithm, harmonic series, Euler-Mascheroni constant
Eric W. Weisstein (MathWorld)
Logarithm, Common Logarithm, Natural Logarithm, Natural Logarithm of 2, Cologarithm,
Alternating Harmonic Series, Harmonic Series
Draw Function Graphs
Logarithm, Common logarithm, Natural logarithm, Exponential function
(1) F. G. Gauß: four-digit logarithmic and trigonometric tables, Stuttgart 1953
(2) Lambacher / Schweizer: Analysis, Stuttgart 1954
(3) H.v. Mangoldt / K.Knopp: Introduction to Higher Mathematics 2, Leipzig 1957
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© 2012 Jürgen Köller
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