Question

How do you simplify `4lne+lne`?

Answer 1

`4ln e + ln e = color(green)(5)`

Explanation:

`ln color(red)(e)`
`color(white)("XXX")=log_color(blue)(e) color(red)(e)`

Remembering that
`color(white)("XXX")log_color(blue)(b) color(red)(a)=c`
means
`color(white)("XXX")color(blue)(b)^c=color(red)(a)`

Therefore
`color(white)("XXX")log_color(blue)(e)color(red)(e)=c`
means
`color(white)("XXX")color(blue)(e)^c=color(red)(e)`

That is
`color(white)("XXX")color(green)(ln_e)=log_e e (= c) = color(green)(1)`

Therefore
`color(white)("XXX")4color(green)(ln_e) + color(green)(ln_e)`

`color(white)("XXX")=4(color(green)(1))+ (color(green)(1))`

`color(white)("XXX")=5`