Solution 3.4:3c
From Förberedande kurs i matematik 1
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- | {{ | + | 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>}} |
- | { | + | |
- | < | + | 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). | ||
+ | |||
+ | The equation rewritten in this way is, in turn, only satisfied if the arguments | ||
+ | <math>x(x+4)</math> and <math>2x+3</math> are equal to each other and positive, i.e. | ||
+ | |||
+ | {{Displayed math||<math>x(x+4) = 2x+3\,\textrm{.}</math>}} | ||
+ | |||
+ | We rewrite this equation as <math>x^2+2x-3=0</math> and completing the square gives | ||
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+ | {{Displayed math||<math>\begin{align} | ||
+ | (x+1)^2-1^2-3 &= 0\,,\\ | ||
+ | (x+1)^2=4\,, | ||
+ | \end{align}</math>}} | ||
+ | |||
+ | which means that <math>x=-1\pm 2</math>, i.e. <math>x=-3</math> and <math>x=1\,</math>. | ||
+ | |||
+ | 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>. |
Current revision
With the log laws, we can write the left-hand side as one logarithmic expression,
\displaystyle \ln x+\ln (x+4) = \ln (x(x+4))\,, |
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
\displaystyle \ln (x(x+4)) = \ln (2x+3) |
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.
\displaystyle x(x+4) = 2x+3\,\textrm{.} |
We rewrite this equation as \displaystyle x^2+2x-3=0 and completing the square gives
\displaystyle \begin{align}
(x+1)^2-1^2-3 &= 0\,,\\ (x+1)^2=4\,, \end{align} |
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\,.