How do solve $3e^x=2e-x+4$ ?

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Answer 1 · 2015-11-08T21:09:44

If you mean $3e^x=2e^(-x)+4$ , then rearrange as a quadratic in $e^x$ , solve and take logs to find:

$x = ln((2+sqrt(10))/3)$

Assuming you meant $3e^x=2e^(-x)+4$ , first multiply by $e^x$ to get:

$3(e^x)^2 = 2+4(e^x)$

Subtract the right hand side from the left to get:

$3(e^x)^2-4(e^x)-2 = 0$

$e^x = (4+-sqrt(4^2-(4xx3xx-2)))/(2*3) =(4+-sqrt(16+24))/6$

$=(4+=sqrt(40))/6 = (4+-2sqrt(10))/6 = (2+-sqrt(10))/3$

Now $sqrt(10) > 2$ , so to get a positive value for $e^x$ (and hence a Real value for $x$ ) we need to choose the $+$ sign here to find:

$e^x = (2+sqrt(10))/3$

and hence:

$x = ln((2+sqrt(10))/3)$

How do solve 3e^x=2e-x+43e^x=2e-x+4?