Question

How do you condense `1/ 7 ln(x + 2)^7 + 1/ 2 ln x − ln(x^2 + 3x + 2)^2`?

Answer 1

`ln((x^(1/2)(x+2))/(x^2+3x+2)^2)`

Explanation:

Using the `color(blue)"laws of logarithms"`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(a/a)color(black)(logx^nhArrnlogx)color(white)(a/a)|)))`

`rArr1/7ln(x+2)^7=ln(x+2)^(7xx1/7)=ln(x+2)`

and `1/2lnx=lnx^(1/2)`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(a/a)color(black)(logx+logy=log(xy))color(white)(a/a)|)))`

`rArrln(x+2)+lnx^(1/2)=lnx^(1/2)(x+2)`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(a/a)color(black)(logx-logy=log(x/y))color(white)(a/a)|)))`

`rArrlnx^(1/2)(x+2)-ln(x^2+3x+2)^2=ln((x^(1/2)(x+2))/(x^2+3x+2)^2)`