Question #0be98

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Question #0be98

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Answer 1 · 2017-11-08T15:19:17

$x \approx 3.53 or x \approx - 4.53$ $x~~3.53 or x~~-4.53$

Explanation:

$log (x + 3) + log (x - 2) = 1$ $log(x+3)+log(x-2)=1$

Using the addition rule, we get:
$log (x^{2} + x - 6) = 1$ $log(x^2+x-6)=1$

$10^{log (x^{2} + x - 6)} = 10^{1} = 10$ $10^(log(x^2+x-6))=10^1=10$

$x^{2} + x - 6 = 10$ $x^2+x-6=10$

$x^{2} + x - 16 = 0$ $x^2+x-16=0$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

$x = \frac{- 1 - \sqrt{1^{2} - 4 (- 16)}}{2} = \frac{- 1 - \sqrt{65}}{2} \approx 3.53$ $x=(-1-sqrt(1^2-4(-16)))/2=(-1-sqrt(65))/2~~3.53$

$x = \frac{- 1 + \sqrt{1^{2} - 4 (- 16)}}{2} = \frac{- 1 + \sqrt{65}}{2} \approx - 4.53$ $x=(-1+sqrt(1^2-4(-16)))/2=(-1+sqrt(65))/2~~-4.53$

Answer 2 · 2017-11-18T12:45:52

$x = - \frac{1}{2} + \frac{\sqrt{65}}{2} \approx 3.53$ $x=-1/2+sqrt65/2~~3.53$

Explanation:

$log (x + 3) + log (x - 2) = 1$ $log(x+3)+log(x-2)= 1$
$log [(x + 3) (x - 2)] = 1$ $log[(x + 3)(x-2)]=1$
$x^{2} + x - 6 = 10$ $x^2+x-6=10$
$x^{2} + x + {(\frac{1}{2})}^{2} = 10 + 6 + \frac{1}{4}$ $x^2+x+(1/2)^2=10+6+1/4$
${(x + \frac{1}{2})}^{2} = \frac{65}{4}$ $(x+1/2)^2= 65/4$
$x + \frac{1}{2} = \pm \frac{\sqrt{65}}{2}$ $x+1/2=+-sqrt65/2$
$x = - \frac{1}{2} \pm \frac{\sqrt{65}}{2}$ $x=-1/2+-sqrt65/2$
$x \approx - 4.53, 3.53$ $x~~ -4.53, 3.53$ => reject the negative root, since it's not in the domain and makes the equation undefined thus it's an "Extraneous"
root.
therefore:
$x \approx 3.53$ $x~~3.53$ => the only valid solution

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