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

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

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Answer 1 · 2015-02-14T19:26:22

You can use the property of the log:
${log}_{a} M - {log}_{a} N = {log}_{a} (\frac{M}{N})$ $log_aM-log_aN=log_a(M/N)$
So you have:
${log}_{6} (\frac{3 x}{x + 1}) = {log}_{6} (1)$ $log_6((3x)/(x+1))=log_6(1)$
${log}_{6} (\frac{3 x}{x + 1}) = 0$ $log_6((3x)/(x+1))=0$
Because from the definition od logarithm:
${log}_{a} b = x \Rightarrow a^{x} = b$ $log_ab=x => a^x=b$
and: ${log}_{6} (1) = 0$ $log_6(1)=0$

as for:
${log}_{6} (\frac{3 x}{x + 1}) = 0$ $log_6((3x)/(x+1))=0$
again you get:
$(\frac{3 x}{x + 1}) = 6^{0} = 1$ $((3x)/(x+1))=6^0=1$
$3 x = x + 1$ $3x=x+1$
$x = \frac{1}{2}$ $x=1/2$

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

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