Question #04717

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Answer 1 · 2017-03-23T13:16:01

Given: $ln((x-5)/(x-1))=ln(6)$

Subtract $ln(6)$ from both sides:

$ln((x-5)/(x-1))-ln(6)= 0$

Use the property of logarithms where subtracting is equivalent to division within the argument:

$ln((x-5)/(6(x-1)))= 0$

Make both sides the exponent of e:

$e^(ln((x-5)/(6(x-1))))= e^0$

The logarithm and the exponential on the left side cancel and the right side becomes 1:

$(x-5)/(6(x-1))= 1$

Now, it is easy to solve for x:

$x-5=6x-6$

$5x=1$

$x = 1/5$

Check:

$ln((1/5-5)/(1/5-1))=ln(6)$

$ln((1-25)/(1-5))=ln(6)$

$ln((-24)/(-4))=ln(6)$

$ln(6) = ln(6)$

This checks.

$x = 1/5$

Answer 2 · 2017-03-24T00:20:43

$x=1/5$

The contents of the two logarithms, which are the only things present on either side of the equation, are equal:

$(x-5)/(x-1)=6$

$=>x-5=6(x-1)$

$=>x-5=6x-6$

$=>1=5x$

$=>x=1/5$