How do you solve ${log}_{6} x + {log}_{6} (x - 9) = 2$ $log_6x+log_6(x-9)=2$ ?

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Answer 1 · 2016-07-24T03:31:30

Use the log property ${log}_{a} (n) + {log}_{a} (m) = {log}_{a} (n \times m)$ $log_a(n) + log_a(m) = log_a(n xx m)$ :

$\Rightarrow {log}_{6} (x (x - 9)) = 2$ $=>log_6(x(x - 9)) = 2$

$\Rightarrow {log}_{6} (x^{2} - 9 x) = 2$ $=>log_6(x^2 - 9x) = 2$

$\Rightarrow x^{2} - 9 x = 6^{2}$ $=>x^2 - 9x = 6^2$

$\Rightarrow x^{2} - 9 x = 36$ $=>x^2 - 9x = 36$

$\Rightarrow x^{2} - 9 x - 36 = 0$ $=>x^2 - 9x - 36 = 0$

$\Rightarrow (x - 12) (x + 3) = 0$ $=>(x - 12)(x + 3) = 0$

$\Rightarrow x = 12 and - 3$ $=>x = 12 and -3$

Checking in the original equation, we find that only $x = 12$ $x = 12$ works. Hence, the solution set is ${12}$ ${12}$ .

Hopefully this helps!

How do you solve log_6x+log_6(x-9)=2log6x+log6(x−9)=2log_6x+log_6(x-9)=2?