Making up of the logarithmic equations by complication method

Составление логарифмических уравнений методом осложнений

Ф.И. Сейтгалиева

Гимназия №2 им. Г. Каирбекова, г. Астана, Казахстан

log3 (10x2 + 5x) = log3 45x.

Let's notice that the complicated equation log ₃ (10 x ² + 5 x ) = log₃ 45 x has the same decision, as the standard equation for in the course of the decision added a difference of identical expressions.

Task    2.     log3(2 x +1) = 2          ^

log3 f (x) = log c ф(x).

Decision. It is necessary to make change to structure of the requirement so that the basis of a logarithm was another. Let's take for the logarithm basis number 9 and we will add a difference log₉ 5 x — log₉ 5 x , thereby complicating the equations. Thus process of the decision will look as follows:

log3 (2x +1) + log9 5x — log9 5x = 2,

Let's notice that mathematics teachers as a whole rather well are able to state ready knowledge, achieving thus quite good results in assimilation by school students of system of special knowledge, that is process of teaching of mathematics is carried out now successfully. However other component of process of training – the doctrine of school students is shown in practice more not enough.

According to David K. Pugalee [2], Karl Wesley Kosko and Anderson Norton [1] in the course of the solution of tasks the difference of identical expressions

log3 (2x +1) + log9 5x — logg 5x = 2,

log3 (2x +1) + log3 5 x = 2 + log9 5x, log39

log3(2 x +1) + ^

= log9 81 + log9 5x,

log3 V5x(2x +1) = log9 (81 • 5x),

log3 V5x (2x +1) = log9 405x.

Solution of a task: log₃ (2 x + 1) = 2      ^

2. Pugalee D.K. Writing, mathematics and metacognition: looking for connections through students' work in mathematical problem solving // School science and mathematics. – 2010. – Vol. 101, № 5. – Р. 236–245.

log₃ V5 x (2 x + 1) = log₉ 405 x.

Task    3.      log₃(2 x + 1) = 2           ^

(2 x + 1)^{log 3} " ^{( x} ) = f ( x ).

Decision. By equation drawing up as initial object we take the equation, the left part of a sign of follow-ing« ^ », and in the right part - for a reference point. So,

H :log₃(2 x + 1) = 2 ,

O : (2 x + 1)^{log 3 ф ( x} ) = f ( x ).

In About such information contains: it is necessary to pass to indicative function with the basis 2х+1, and with preservation of logarithmic function log3(^), only having changed its expression. This requirement we can execute when we will use that log3 (•) log3 (2x +1) = log3 (2x + 1)log3(,).

So, partially nature of change of structure already was defined. But we should keep an equal-sign. This condition will be executed when both parts H we will increase log a ( ^e ) / log₃( » ) .

In structure of a reference point concrete functions are not given. Means instead of a point it is possible to take any function, for example 7х. Then

(log3 7x / log3 7x) log3 (2x +1) = 2, log3(2 x +1)

log₃ 7 x --³-------- = 2,

3       log 7x       , log3(2 x + 1)log37x =2

log 7 x

3 log 3 (2 x + 1)^log3 7 ^x = 3 log 3 49 x ²

(2 x + 1)^{log 3} 7 x = 49 x ².

So, the complicated equation is received:

log₃(2 x + 1) = 2 ^ (2 x + 1)^{log 3} 7 ^x = 49 x ².

For ensuring understanding of process of transformation of this equation considered separately a case of addition of zero and multiplication to unit. In general, introduction of zero and unit can be carried out in any place of transformation. Now, more than ever, becomes clear that the mathematics is not only set of the facts stated in the form of theorems, but first of all – the arsenal of methods and even that is still before language for the description of the facts and methods of the most different areas of a science and practical activities.