Lösung 3.4:3c
Aus Online Mathematik Brückenkurs 1
K |
|||
| Zeile 1: | Zeile 1: | ||
With the log laws, we can write the left-hand side as one logarithmic expression, | With the log laws, we can write the left-hand side as one logarithmic expression, | ||
| + | {{Displayed math||<math>\ln x+\ln (x+4) = \ln (x(x+4))\,,</math>}} | ||
| - | <math>\ln x | + | but this rewriting presupposes that the expressions <math>\ln x</math> and <math>\ln (x+4)</math> are defined, i.e. <math>x > 0</math> and <math>x+4 > 0\,</math>. Therefore, if we choose to continue with the equation |
| + | {{Displayed math||<math>\ln (x(x+4)) = \ln (2x+3)</math>}} | ||
| - | + | we must remember to permit only solutions that satisfy <math>x > 0</math> (the condition <math>x+\text{4}>0</math> is then automatically satisfied). | |
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | we must remember to permit only solutions that satisfy | + | |
| - | <math>x>0</math> | + | |
| - | (the condition | + | |
| - | <math>x+\text{4}>0</math> | + | |
| - | is then automatically satisfied). | + | |
The equation rewritten in this way is, in turn, only satisfied if the arguments | The equation rewritten in this way is, in turn, only satisfied if the arguments | ||
| - | <math>x | + | <math>x(x+4)</math> and <math>2x+3</math> are equal to each other and positive, i.e. |
| - | and | + | |
| - | <math> | + | |
| - | are equal to each other and positive, i.e. | + | |
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| + | {{Displayed math||<math>x(x+4) = 2x+3\,\textrm{.}</math>}} | ||
| - | <math> | + | We rewrite this equation as <math>x^2+2x-3=0</math> and completing the square gives |
| - | + | ||
| - | + | ||
| - | + | ||
| + | {{Displayed math||<math>\begin{align} | ||
| + | (x+1)^2-1^2-3 &= 0\,,\\ | ||
| + | (x+1)^2=4\,, | ||
| + | \end{align}</math>}} | ||
| - | which means that | + | which means that <math>x=-1\pm 2</math>, i.e. <math>x=-3</math> and <math>x=1\,</math>. |
| - | <math>x=- | + | |
| - | <math>x=- | + | |
| - | and | + | |
| - | <math>x= | + | |
| - | Because | + | Because <math>x=-3</math> is negative, we neglect it, whilst for <math>x=1</math> we have both that <math>x > 0</math> and <math>x(x+4) = 2x+3 > 0\,</math>. Therefore, the answer is <math>x=1\,</math>. |
| - | <math>x=- | + | |
| - | is negative, we neglect it, whilst for | + | |
| - | <math>x= | + | |
| - | we have both that | + | |
| - | <math>x>0 | + | |
| - | and | + | |
| - | <math>x | + | |
| - | <math>x= | + | |
Version vom 14:34, 2. Okt. 2008
With the log laws, we can write the left-hand side as one logarithmic expression,
but this rewriting presupposes that the expressions \displaystyle \ln x and \displaystyle \ln (x+4) are defined, i.e. \displaystyle x > 0 and \displaystyle x+4 > 0\,. Therefore, if we choose to continue with the equation
we must remember to permit only solutions that satisfy \displaystyle x > 0 (the condition \displaystyle x+\text{4}>0 is then automatically satisfied).
The equation rewritten in this way is, in turn, only satisfied if the arguments \displaystyle x(x+4) and \displaystyle 2x+3 are equal to each other and positive, i.e.
We rewrite this equation as \displaystyle x^2+2x-3=0 and completing the square gives
which means that \displaystyle x=-1\pm 2, i.e. \displaystyle x=-3 and \displaystyle x=1\,.
Because \displaystyle x=-3 is negative, we neglect it, whilst for \displaystyle x=1 we have both that \displaystyle x > 0 and \displaystyle x(x+4) = 2x+3 > 0\,. Therefore, the answer is \displaystyle x=1\,.
