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Mathematics 7 Online
OpenStudy (anonymous):

solve equation e^x + 24e^−x − 10 = 0

OpenStudy (amity):

2 solutions one is x=ln6\2 other is ln4\2

OpenStudy (anonymous):

@amity how?

OpenStudy (tkhunny):

Multiply be \(e^{x}\), which never is 0, and it should take on the appearance of something quadratic.

OpenStudy (amity):

Let e^x=y,then equation is y+24/y=10,,,,that is y^2-10y+24=0

OpenStudy (amity):

Let e^x=y,then equation is y+24/y=10,,,,that is y^2-10y+24=0

OpenStudy (anonymous):

I still don't understand this question. Am I supposed to get two answers?

OpenStudy (amity):

Factor is y^2-6y-4y+24=0,,,y(y-6)-4(y-6)=0,,,(y-4)(y-6)=0,,..which yields y=6, y=4,,,e^x is 4 and 6

OpenStudy (amity):

Factor is y^2-6y-4y+24=0,,,y(y-6)-4(y-6)=0,,,(y-4)(y-6)=0,,..which yields y=6, y=4,,,e^x is 4 and 6

OpenStudy (triciaal):

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