Question #fc511

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Answer 1 · 2017-02-17T22:20:57

We have that

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

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

$log [(2 x + 1) (x - 2)] = 1$ $log[(2x+1)(x-2)]=1$ [Use the additive law of logarithms]

$log [(2 x + 1) (x - 2)] = log 10$ $log[(2x+1)(x-2)]=log10$ [We used the base 10 logarithm here]

$(2 x + 1) (x - 2) = 10$ $(2x+1)(x-2)=10$

$2 x^{2} - 3 x - 12 = 0$ $2 x^2 - 3 x - 12 = 0$

The last one (quadratic equation) has solutions

$x_{1} = \frac{3}{4} - \frac{\sqrt{105}}{4}$ $x_1 = 3/4 - sqrt(105)/4$ or $x_{1} \approx - 1.8117$ $x_1≈-1.8117$

$x_{2} = \frac{3}{4} + \frac{\sqrt{105}}{4}$ $x_2= 3/4 + sqrt(105)/4$ or $x_{2} \approx 3.3117$ $x_2≈3.3117$

Because $(x - 2) \geq 0$ $(x-2)>=0$ hence $x \geq 2$ $x>=2$ the acceptable solution is

$x_{2} = \frac{3}{4} + \frac{\sqrt{105}}{4}$ $x_2= 3/4 + sqrt(105)/4$