How do you solve ${log}_{4} (x - 7) + {log}_{4} (x + 7) = 2$ $log_4 (x - 7) + log_4 (x + 7) = 2$ ?

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How do you solve ${log}_{4} (x - 7) + {log}_{4} (x + 7) = 2$ $log_4 (x - 7) + log_4 (x + 7) = 2$ ?

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Answer 1 · 2016-04-23T15:51:19

$x = + \sqrt{65}$ $x=+sqrt65$

Explanation:

$\Rightarrow {log}_{4} ((x - 7) \cdot (x + 7)) = 2$ $=>log_4((x-7)*(x+7))=2$
$\Rightarrow x^{2} - 7^{2} = 4^{2}$ $=>x^2-7^2=4^2$
$\Rightarrow x^{2} = 49 + 16 = 65$ $=>x^2=49+16=65$
$x = \pm \sqrt{65}$ $x=+-sqrt65$
negative root not possible as log of negative number is not defined
so $x = \sqrt{65}$ $x=sqrt65$

Answer 2 · 2016-04-23T16:09:48

$\sqrt{65} \approx 8.062$ $sqrt 65~~8.062$

Explanation:

${log}_{b} m + {log}_{b} n = {log}_{b} (m n) .$ $log_b m+log_b n=log_b (mn).$
We can use this here and condense, assuming that x > 7..

${log}_{4} ((x - 7) (x + 7)) = 2$ $log_4((x-7)(x+7))=2$

Inversely, $(x - 7) \cdot (x + 7) = 4^{2}$ $(x-7)*(x+7)=4^2$ .

$x^{2} - 49 = 16$ $x^2-49=16$

$x = \pm \sqrt{65}$ $x=+-sqrt65$

Negative root is inadmissible, as x > 7.

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How do you solve ${log}_{4} (x - 7) + {log}_{4} (x + 7) = 2$ $log_4 (x - 7) + log_4 (x + 7) = 2$ ?

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Explanation:

Explanation:

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