Math help

ex - 6e-x = 1

$e^x-6e^{-x}=1\\\\ e^x-\frac{6}{e^{x}}=1\\\\ e^x(e^x-\frac{6}{e^{x}})=1*e^x\\\\ (e^x)^2-6=e^x\\\\ (e^x)^2-e^x-6=0\\\\ sub\;y=e^x\\\\ y^2-y-6=0\\\\ (y-3)(y+2)=0 y=3\;\;or\;\;y=-2\\\\ e^x=3\;\;or\;\;e^x=-2\\\\ lne^x=ln3\;\;or\;\;no\; solution\\\\ x=ln3$

e^x - 6e^-x = 1      multiply through by e^x

e^2x - 6 = e^x      subtract  ex  from both sides and rearrange the left side

e^2x - e^x - 6 =  0     factor this

(e^x - 3) (e^x + 2) =     setting each factor to 0, we have

e^x - 3 = 0                                   and                             e^x + 2 = 0

The second equation has no solution.....for the first, we have

e^x - 3 = 0

Add 3 to both sides

e^x  = 3       take the ln of both sides

ln e^x  = ln 3      and we can write

x ln e =   ln 3     and ln e = 1, so we can ignore this.......and we have

x = ln 3 = about 1.0986

HAHAHAHAHA!!!!!!!!.......beat you, Melody   !!!!!!    LOL!!!!!!!