How do you solve ${log}_{6} (x + 3) = 1 - {log}_{6} (x - 2)$ $Log_6 (x+3) = 1- log_6 (x-2)$ ?

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Answer 1 · 2018-06-26T10:03:56

$x = 3$ $x=3$

${log}_{6} (x + 3) = 1 + {log}_{6} (x - 2)$ $log_6(x+3)=1+log_6(x-2)$

${log}_{6} (x + 3) = {log}_{6} 6 + {log}_{6} (x - 2)$ $log_6(x+3)=log_6 6+log_6(x-2)$

${log}_{6} (x + 3) = {log}_{6} 6 (x - 2)$ $log_6(x+3)=log_6 6(x-2)$

$x + 3 = 6 (x - 2)$ $x+3=6(x-2)$

$x + 3 = 6 x - 12$ $x+3=6x-12$

$5 x = 15$ $5x=15$

$x = 3$ $x=3$

Remember:
${log}_{a} b + {log}_{a} c = {log}_{a} b c$ $log_a b+log_a c=log_a bc$
${log}_{a} a = 1$ $log_a a=1$

How do you solve log6(x+3)=1−log6(x−2)Log_6 (x+3) = 1- log_6 (x-2)?