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)#