How do you solve ${log}_{3} (x + 1) = 2 + {log}_{3} (x - 3)$ $Log_3 (x+1) = 2 + Log_3 (x-3)$ ?

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Answer 1 · 2016-05-04T18:34:02

$x = \frac{7}{2}$ $x=7/2$

${log}_{3} (x + 1) - {log}_{3} (x - 3) = 2$ $log_3 (x+1)-log_3(x-3)=2$

${log}_{3} (\frac{x + 1}{x - 3}) = 2$ $log_3 ((x+1)/(x-3))=2$

$3^{2} = \frac{x + 1}{x - 3}$ $3^2 = (x+1)/(x-3)$

$9 = \frac{x + 1}{x - 3}$ $9=(x+1)/(x-3)$

$9 (x - 3) = x + 1$ $9(x-3)=x+1$

$9 x - 27 = x + 1$ $9x-27=x+1$

$9 x - x = 1 + 27$ $9x-x=1+27$

$8 x = 28$ $8x=28$

$x = \frac{28}{8} = \frac{7}{2}$ $x=28/8=7/2$

How do you solve log3(x+1)=2+log3(x−3)Log_3 (x+1) = 2 + Log_3 (x-3)?