How do you solve $ln (x - 6) - ln (5) = ln (7) + ln (x - 2)$ $ln(x-6) - ln(5) = ln(7) + ln(x-2)$ ?

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How do you solve $ln (x - 6) - ln (5) = ln (7) + ln (x - 2)$ $ln(x-6) - ln(5) = ln(7) + ln(x-2)$ ?

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Answer 1 · 2016-07-10T09:01:41

$x = \frac{65}{34}$ $x=65/34$

Explanation:

Subtraction of logs is the consequence of the source values being divided. So $ln (x - 6) - ln (5) \to ln (\frac{x - 5}{5})$ $ln(x-6)-ln(5) ->ln((x-5)/5)$

Addition of logs is the consequence of the source values being multiplied. So $ln (7) + ln (x - 2) \to ln (7 (x - 2))$ $ln(7)+ln(x-2)->ln(7(x-2))$

Putting this all together

$ln (\frac{x - 5}{5}) = ln (7 (x - 2))$ $ln((x-5)/5)=ln(7(x-2))$

$\Rightarrow \frac{x - 5}{5} = 7 (x - 2)$ $=>(x-5)/5=7(x-2)$

$x - 5 = 35 (x - 2)$ $x-5=35(x-2)$

$x - 5 = 35 x - 70$ $x-5=35x-70$

$34 x = 65$ $34x=65$

$x = \frac{65}{34}$ $x=65/34$

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How do you solve $ln (x - 6) - ln (5) = ln (7) + ln (x - 2)$ $ln(x-6) - ln(5) = ln(7) + ln(x-2)$ ?

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Explanation:

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How do you solve ln(x−6)−ln(5)=ln(7)+ln(x−2)ln(x-6) - ln(5) = ln(7) + ln(x-2)?

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How do you solve $ln (x - 6) - ln (5) = ln (7) + ln (x - 2)$ $ln(x-6) - ln(5) = ln(7) + ln(x-2)$ ?