How do you solve ${log}_{b} (x - 1) + {log}_{b} 3 = {log}_{b} x$ $log_b(x-1)+log_b3=log_bx$ ?

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Answer 1 · 2016-07-14T01:37:59

I'm assuming you want to solve for x...

${log}_{b} (3) = {log}_{b} (x) - {log}_{b} (x - 1)$ $log_b(3) = log_b(x) - log_b(x - 1)$

${log}_{b} (3) = {log}_{b} (\frac{x}{x - 1})$ $log_b(3) = log_b((x)/(x - 1))$

Since we're in equal bases we can eliminate:

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

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

$3 x - 3 = x$ $3x - 3 = x$

$2 x = 3$ $2x = 3$

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

Hopefully this helps!

How do you solve logb(x−1)+logb3=logbxlog_b(x-1)+log_b3=log_bx?