burhan101
Simplify this expression



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burhan101
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\[ \huge \ln e^{\sin^2x} + \ln e^{\cos^2x} \]

carlos74923
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hah.,
a

burhan101
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the answer is one.. but how ?!

carlos74923
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its a nice problem

carlos74923
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its 0

burhan101
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no my answerkey says 1

carlos74923
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:'( walks in shame

chrissy401
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\[\cos^2x \ln e+\sin^2x \ln e\]
but \[\ln e\] is 1

chrissy401
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since \[\ln e = \log_{e} ^e\]

burhan101
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\[\sin^2x+\cos^2x\]

chrissy401
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we know that \[\cos^2x+\sin^2x=1\]
in trigonometry

chrissy401
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thast why the answer is one

carlos74923
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log e is = to lin

burhan101
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thank you :D

carlos74923
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so log e to e to 1 is 1

chrissy401
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you're welcome..