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fan1597534862
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Find, as natural logarithms, the root of the equation

2e^(x) + 3e^(-x) = 7
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SH0405
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(Original post by fan1597534862)
Find, as natural logarithms, the root of the equation

2e^(x) + 3e^(-x) = 7
e^(-x)[2e^(2x) + 3] = 7

--> 2e^(2x) + 3 = 7*(1/e^(-x))
= 7e^x

--> 2e^(2x) - 7e^x + 3 = 0

set A=e^x

--> 2A^2 - 7A + 3 = 0
--> 2A^2 - 6A - A + 3 = 0
--> (2A-1)(A-3) = 0 --> A=½ & A=3

--> e^x=½ --> x = ln(½) & e^x=3 --> x = ln3

Hope this helps.
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