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Mathematics 10 Online
OpenStudy (anonymous):

how do i simplify In2x+In3-Inx

OpenStudy (cwrw238):

use the laws of logarithms ln x + ln y = ln xy lnx - ln y = ln x/y so ln2x + ln3 = ln (3*2x) = ln 6x so now you need to simplify ln 6x - ln x

OpenStudy (anonymous):

thank you very much

OpenStudy (cwrw238):

yw

OpenStudy (anonymous):

the answer in the book is just given as 6 can you explain that

OpenStudy (cwrw238):

by the second of the laws in my first post: ln 6x - ln x - ln (6x /x) = ln 6

OpenStudy (anonymous):

thanks again

OpenStudy (cwrw238):

* = ln (6x/x) = ln 6

OpenStudy (anonymous):

what about the laws for say In(x^3)+In(x^2)-Inx

OpenStudy (cwrw238):

- use same method as before apply the first law then second

OpenStudy (anonymous):

im working on it

OpenStudy (anonymous):

is it right to say that Ina-Inb = Ina/b

OpenStudy (anonymous):

can you confirm i have used the correct logic In(x^3)/In(x^2) =Inx so Inx/Inx = 0

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