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

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

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Answer 1 · 2016-08-17T23:48:10

$x = \frac{3}{2}$ $x = 3/2$

Explanation:

${log}_{3} x - {log}_{3} (x - 1) = 1 = {log}_{3} 3$ $log_3 x-log_3(x-1)=1 = log_3 3$ so
${log}_{3} (\frac{x}{x - 1}) = {log}_{3} 3$ $log_3(x/(x-1))=log_3 3$ so
$\frac{x}{x - 1} = 3$ $x/(x-1) = 3$ or
$x = 3 (x - 1)$ $x =3(x-1)$ Solving for $x$ $x$ gives

$x = \frac{3}{2}$ $x = 3/2$ . Verifying the feasibility

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

So, $x = \frac{3}{2}$ $x = 3/2$ is the solution.

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

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Explanation:

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