How do you solve ${log}_{3} x + {log}_{3} (x - 8) = 2$ $log_3x+log_3(x-8)=2$ ?

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Answer 1 · 2018-03-31T13:54:27

Thus, x=9 is the valid solution

${log}_{3} x + {log}_{3} (x - 8) = 2$ $log_3 x+log_3 (x-8)=2$

$log m + log n = log m n$ $logm+logn=logmn$

${log}_{3} x + {log}_{3} (x - 8) = {log}_{3} x (x - 8)$ $log_3 x+log_3 (x-8)=log_3 x(x-8)$

${log}_{3} x (x - 8) = 2$ $log_3 x(x-8)=2$

$x (x - 8) = 3^{2}$ $x(x-8)=3^2$

$x (x - 8) = 9$ $x(x-8)=9$

$x^{2} - 8 x = 9$ $x^2-8x=9$

$x^{2} - 8 x - 9 = 0$ $x^2-8x-9=0$

$x^{2} - 9 x + 1 x - 9 = 0$ $x^2-9x+1x-9=0$

$x (x - 9) + 1 (x - 9) = 0$ $x(x-9)+1(x-9)=0$

$(x + 1) (x - 9) = 0$ $(x+1)(x-9)=0$

$x = - 1, x = 9$ $x=-1, x=9$

$x = - 1, 9$ $x=-1,9$

Here, x needs to be a positive number .

Thus, x=9 is the valid solution

How do you solve log3x+log3(x−8)=2log_3x+log_3(x-8)=2?