How do you solve for x: 6 = #e^x+5e^-x# ?

#e^x+5e^-x#

1 Answer
Apr 2, 2018

The solutions are #x=0# and #x=ln5#.

Explanation:

Multiply everything by #e^x#:

#6=e^x+5e^-x#

#6color(blue)(*e^x)=e^xcolor(blue)(*e^x)+5e^-xcolor(blue)(*e^x)#

#6color(blue)(e^x)=e^(x+color(blue)x)+5e^(-x+color(blue)x)#

#6color(blue)(e^x)=e^(2x)+5e^(0)#

#6color(blue)(e^x)=e^(2x)+5*1#

#6color(blue)(e^x)=e^(2x)+5#

#0=e^(2x)-6color(blue)(e^x)+5#

#0=(color(blue)(e^x))^2-6color(blue)(e^x)+5#

Let #u=e^x#:

#0=u^2-6u+5#

#0=(u-5)(u-1)#

#u=1,5#

Put #e^x# back in for #u#:

#color(white){color(black)( (e^x=1,qquade^x=5), (color(blue)ln(color(black)(e^x))=color(blue)ln(color(black)1),qquadcolor(blue)ln(color(black)(e^x))=color(blue)ln(color(black)5)), (x=0,qquadx=color(blue)ln(color(black)5))):}#

These are the solutions. Hope this helped!