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OpenStudy (anonymous):
use the law on exponents e^(a + b) = e^a * e^b
then use then use the law on logarithms e^(ln c) = c
OpenStudy (anonymous):
HOW?
OpenStudy (anonymous):
\[e^{a+b} = e^a.e^b\]
\[e^{lnx} = x\]
Use the above formulae you will get your answer...
OpenStudy (anonymous):
WOULD you please elaborate more
OpenStudy (anonymous):
What is there in place of a??
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OpenStudy (anonymous):
i dont know
OpenStudy (anonymous):
See,
Compare:
\[e^{1 + lnx} = e^{a + b}\]
Now compare it and tell me what is there in place of a?
OpenStudy (anonymous):
1
OpenStudy (anonymous):
Yes..
And in place of b?
OpenStudy (anonymous):
lnx
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OpenStudy (anonymous):
Yes..
\[e^{a+b}= e^a.e^b\]
Apply this formula and you know a = 1 and b= lnx
OpenStudy (anonymous):
i dont get it on how to do. would you plese solve it for me?
OpenStudy (anonymous):
Why??
Where are you facing problem..
See when the bases are same their exponents if in sum can be brought to multiplication as that formula is saying...
OpenStudy (anonymous):
I DONT see the connection to my question
OpenStudy (anonymous):
\[e^{1 + lnx} = e^1.e^{lnx}\]
What is the problem you are facing in doing this??
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OpenStudy (anonymous):
once you have e^1 x e^(lnx) do you leave it as is or do the bases cancel?
OpenStudy (anonymous):
And you know now:
\[e^{lnx} = x\]
OpenStudy (anonymous):
Oh okay. I can see it now. Sorry the other girl was so rude. Thanks so much!