Question

find an approximate solution to the equation. Round any intermediate work and steps to three decimal places.

Answer 1

\(7ln~(x+2.8)=11.2\)

Answer 2

Remember :-
\[\large \bf \ln(x)=\log_{e} x~~~~where, ~~~e=euler's~constant ≈ 2.71828183\]

Answer 3

where , e =2.718

Answer 4

So,suppose :-
\[\large \bf \ln(x+2.8)=y\]
therefore,we get

Answer 5

\[\large \bf 7y=11.2\]
and find out the value of y first ?

Answer 6

@sleepyjess

Answer 7

y=1.6
sorry it wouldnt load until now so i couldnt see what you said

Answer 8

good,so we got, y=1.6

Answer 9

therefore,
\[\large \bf \ln(x+2.8)=1.6\]
\[\large \bf \log_{e}(x+2.8)=1.6 \]

Answer 10

Remember :-
\[\huge \bf \log_{x} y=n----->x^n=y\]

Answer 11

use this property..

Answer 12

\(e^{1.6}=x+2.8\)

Answer 13

good

Answer 14

and we have , e=2.7

Answer 15

plug in the value !!

Answer 16

what you get ????

Answer 17

4.9=x+2.8
2.1=x

Answer 18

yes.....

Answer 19

therefore,x=2.1

Answer 20

@mayankdevnani Sorry to bother you again... I understand everything except where you said 7y=11.2. How did you get this?

Answer 21

i supposed \[\large \bf \ln(x+2.8)=y\]

Answer 22

Thanks!

Answer 23

welcome :)