Ln simplification

[Yuseph]

Wussup guys,

Please tell me how the author got from one point to the other. Since log a/b = log a - log b i cant figure out in which scenario he ended up with (x + 3) on top. Unless theres a particular law of order to follow. Thanks

 

[R.M.]

-ln(x -2 ) = ln(x-2)^-1. So your problem can be rewritten as ln(x - 2)² + ln(x-2)^-1 + ln(x + 3). Now that each term is written in terms of addition, you can apply the multiplication rule: (x-2)²(x-2)^-1(x+3) = (x-2)²(x+3)/(x-2).

 

[Yuseph]

Thanks for this new method
Sorry about this question

 

[Deleted member 4993]

Wussup guys,

Please tell me how the author got from one point to the other. Since log a/b = log a - log b i cant figure out in which scenario he ended up with (x + 3) on top. Unless theres a particular law of order to follow. Thanks
View attachment 21989

Another way to rewrite the assigned problem:

\(\displaystyle ln(x-2)^2 \ - \ ln(x-2) \ + \ ln(x+3) = 1.6 \) ....................rearranging LHS

\(\displaystyle \left[ln(x-2)^2 + \ ln(x+3)\right] \ - \ ln(x-2) \ = 1.6 \) ...................Apply laws of log multiplication

\(\displaystyle ln\left[(x-2)^2 (x+3)\right] \ - \ ln(x-2) \ = 1.6 \) ..................Apply laws of log division

\(\displaystyle ln\left[\frac{(x-2)^2 * (x+3)}{x-2}\right] \ = 1.6 \).