How do you solve $ln(x-4)-ln(3x-10)=ln (1/x)$ ?

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Answer 1 · 2016-09-07T13:44:57

$x= 5$

Use the rule $log_a(n) - log_a(m) = log_a(n/m)$ .

$=>ln((x - 4)/(3x - 10)) = ln(1/ x)$

Now use the rule $lna = lnb-> a = b$ .

$(x - 4)/(3x - 10) = 1/x$

$x^2 - 4x = 3x - 10$

$x^2 - 7x + 10 = 0$

$(x - 5)(x - 2) = 0$

$x = 5 and 2$

However, $x = 2$ is extraneous, since it renders the equation undefined.

Hopefully this helps!

How do you solve ln(x-4)-ln(3x-10)=ln (1/x) ln(x-4)-ln(3x-10)=ln (1/x)?