How do you solve ${log}_{3} x + {log}_{3} (x - 8) = 2$ $log_3 x + log_3 (x - 8) = 2$ ?

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Answer 1 · 2016-05-08T14:51:25

x = 9

Use ${log}_{b} m + {log}_{b} n = {log}_{b} (m n)$ $log_b m + log_b n = log_b(mn)$

and inverse of $y = {log}_{b} x$ $y=log_bx$ is $x = b^{y}$ $x = b^y$

Here, ${log}_{3} x + {log}_{b} (x - 8) = {log}_{3} (x (x - 8)) = 2$ $log_3 x+log_b(x-8)= log_3(x(x-8))=2$ .

Inversely, $x (x - 8) = 3^{2} = 9$ $x(x-8)=3^2=9$ .

So, $x^{2} - 8 x - 9 = (x + 1) (x - 9) = 0$ $x^2-8x-9= (x+1)(x-9)=0$ .

$x = - 1 and 9$ $x = -1 and 9$ . Negative x is inadmissible for ${log}_{3} x$ $log_3x$ .

How do you solve log3x+log3(x−8)=2log_3 x + log_3 (x - 8) = 2?