How do you solve ${log}_{3} (2 x + 1) = {log}_{3} (x - 4)$ $log_3 (2x + 1) = log_3 (x - 4)$ ?

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How do you solve ${log}_{3} (2 x + 1) = {log}_{3} (x - 4)$ $log_3 (2x + 1) = log_3 (x - 4)$ ?

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Answer 1 · 2016-05-07T20:42:37

We have that

${log}_{3} (2 x + 1) = {log}_{3} (x - 4) \Rightarrow (2 x + 1) = (x - 4) \Rightarrow 2 x - x = - 1 - 4 \Rightarrow x = - 5$ $log_3 (2x + 1) = log_3 (x - 4)=> (2x+1)=(x-4)=>2x-x=-1-4=>x=-5$

But hence it should be $x > 4$ $x>4$ there no real solutions.

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How do you solve ${log}_{3} (2 x + 1) = {log}_{3} (x - 4)$ $log_3 (2x + 1) = log_3 (x - 4)$ ?

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How do you solve ${log}_{3} (2 x + 1) = {log}_{3} (x - 4)$ $log_3 (2x + 1) = log_3 (x - 4)$ ?