# Thread: adding exponential logs

1. ## adding exponential logs

Hi, i would like to ask you for some help about how to add these exponential logs...

If A = 200e^-0.2m
and B = 10.5e^0.3m

what is A + B

seriously, i cant add them... and what i mean is adding the 2 equations... and by the way i dont know how you guys can make the powers shown here so i just typed '^'... sorry...

2. Originally Posted by popsterrockz
Hi, i would like to ask you for some help about how to add these exponential logs...

If A = 200e^-0.2m
and B = 10.5e^0.3m

what is A + B

seriously, i cant add them... and what i mean is adding the 2 equations... and by the way i dont know how you guys can make the powers shown here so i just typed '^'... sorry...

There's no closed form for adding numbers raised to some power. The only thing you can do, and not always the result is nice, is to factor out:

$\displaystyle 200\,e^{-0.2m}+10.5\,e^{0.3m}=e^{-0.2m}\left(200+10.5\,e^{0.5m}\right)$

Tonio

3. ## exponential logs

Originally Posted by popsterrockz
Hi, i would like to ask you for some help about how to add these exponential logs...

If A = 200e^-0.2m
and B = 10.5e^0.3m

what is A + B

seriously, i cant add them... and what i mean is adding the 2 equations... and by the way i dont know how you guys can make the powers shown here so i just typed '^'... sorry...
Hi pops,

I did this as follows
Take the natural log of A and its equality.Simplify

Same procedure ln B.

I get ln A=-1.26m ln B =m

A+B=e^-1.26 +e^m

bjh

4. Originally Posted by bjhopper
Hi pops,

I did this as follows
Take the natural log of A and its equality.Simplify

Same procedure ln B.

I get ln A=-1.26m ln B =m

A+B=e^-1.26 +e^m

bjh

The above is wrong: it is not true that $\displaystyle \log (A+B)=\log A+\log B$ .

Tonio

5. Originally Posted by tonio
The above is wrong: it is not true that $\displaystyle \log (A+B)=\log A+\log B$ .

Tonio
Hello Tonio,

You may find an error in what I did but i am not adding the ln A +ln B. Iam adding the calculated values of A and B.If you would try this approach I would appreciate knowing where I went wrong.

bjh

6. ## adding exponential logs

Originally Posted by tonio
The above is wrong: it is not true that $\displaystyle \log (A+B)=\log A+\log B$ .

Tonio
I did err in the simplification operations.Here is my revised solution.

A=200 e^-.2m
ln A = ln 200 e^-.2m
ln A =ln 200 -.2m ln e
ln A =5.3 -.2m
e^5.3-.2m =A

B=10.5e^.3m
Ln B = ln 10.5 e^.3m
ln B = ln 10.5 + .3m ln e
lnB =2.35+.3m
e^2.35+.3m =B

A+B= e^5.3-.2m + e^2.35 +.3m

bjh

ln A =ln 200-.2m ln e
ln A =5.3-.2m
e^5.3-.2m =A

7. [quote=bjhopper;504191]I did err in the simplification operations.Here is my revised solution.

A=200 e^-.2m
ln A = ln 200 e^-.2m
ln A =ln 200 -.2m ln e
ln A =5.3 -.2m
e^5.3-.2m =A

This must be $\displaystyle A=e^{5.3-0.2m}$

B=10.5e^.3m
Ln B = ln 10.5 e^.3m
ln B = ln 10.5 + .3m ln e
lnB =2.35+.3m
e^2.35+.3m =B

This must be $\displaystyle B=e^{2.35+0.3m}$

A+B= e^5.3-.2m + e^2.35 +.3m

It must be $\displaystyle A+B=e^{5.3-0.2m}+e^{2.35+0.3m}$ , which is close to what I wrote in my first post (I just didn't do approximations)

Tonio

bjh

8. Originally Posted by bjhopper
Hello Tonio,

You may find an error in what I did but i am not adding the ln A +ln B. Iam adding the calculated values of A and B.If you would try this approach I would appreciate knowing where I went wrong.

True, you didn't do that. I got confused because your calculations are pretty wrong, but you already fixed that in other post.

Tonio

bjh
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