How do you solve ${log}_{4} (x^{2} + 3 x) - {log}_{4} (x - 5) = 1$ $log_4 (x^2 + 3x) - log_4(x - 5) =1$ ?

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How do you solve ${log}_{4} (x^{2} + 3 x) - {log}_{4} (x - 5) = 1$ $log_4 (x^2 + 3x) - log_4(x - 5) =1$ ?

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Answer 1 · 2016-07-28T17:42:11

$x = \frac{1 + \sqrt{79} i}{2}, \frac{1 - \sqrt{79} i}{2}$ $x=(1+sqrt(79)i)/2, (1-sqrt(79)i)/2$

Explanation:

${log}_{4} (x^{2} + 3 x) - {log}_{4} (x - 5) = 1$ $log_4(x^2+3x) - log_4(x-5) = 1$
Using the identity ${log}_{a} b - {log}_{a} c = {log}_{a} (\frac{b}{c})$ $log_ab -log_ac = log_a(b/c)$ , we can simplify the expression to
${log}_{4} (\frac{x^{2} + 3 x}{x - 5}) = 1$ $log_4((x^2+3x)/(x-5)) = 1$
This yields to
$\frac{x^{2} + 3 x}{x - 5} = 4^{1}$ $(x^2+3x)/(x-5) = 4^1$
This is because, if ${log}_{a} b = c$ $log_ab = c$ , then $b = a^{c}$ $b=a^c$
Now the equation can be solved for values of $x$ $x$
$\Rightarrow x^{2} + 3 x = 4^{1} (x - 5)$ $=>x^2+3x = 4^1(x-5)$
$\Rightarrow x^{2} + 3 x = 4 x - 20$ $=> x^2+3x = 4x -20$
$\Rightarrow x^{2} + 3 x - 4 x + 20 = 0$ $=> x^2 + 3x -4x +20 = 0$
$\Rightarrow x^{2} - x + 20 = 0$ $=> x^2-x+20=0$
The roots of this equation are imaginary.
They can be found using the formula $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +- sqrt(b^2-4ac))/(2a)$
Here, $a = 1$ $a=1$ , $b = - 1$ $b=-1$ and $c = 20$ $c=20$
Therefore, $x = \frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot 1 \cdot 20}}{2 \cdot 1}$ $x= (1+- sqrt((-1)^2-4*1*20))/(2*1)$
$\Rightarrow x = \frac{1 + \sqrt{79} i}{2}, \frac{1 - \sqrt{79} i}{2}$ $=>x = (1+sqrt(79)i)/2, (1-sqrt(79)i)/2$

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How do you solve ${log}_{4} (x^{2} + 3 x) - {log}_{4} (x - 5) = 1$ $log_4 (x^2 + 3x) - log_4(x - 5) =1$ ?

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How do you solve log4(x2+3x)−log4(x−5)=1log_4 (x^2 + 3x) - log_4(x - 5) =1?

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How do you solve ${log}_{4} (x^{2} + 3 x) - {log}_{4} (x - 5) = 1$ $log_4 (x^2 + 3x) - log_4(x - 5) =1$ ?