e^-3.5555x = 0

[leahdm02]

e^3.5555x=0
This is a questions on my exponents assignment for functions. It says to solve for x. I have my exam tomorrow , so if anyone knows the answer and could explain it'd be a big help! Thanks! Btw the x is part of the exponent (e to the power of (-3.5555x) equals 0)

 

[pka]

e^3.5555x=0
This is a questions on my exponents assignment for functions. It says to solve for x. I have my exam tomorrow , so if anyone knows the answer and could explain it'd be a big help! Thanks! Btw the x is part of the exponent (e to the power of (-3.5555x) equals 0)

\(\displaystyle (\forall t)[e^t\ne 0]\) BTW \(\displaystyle (\forall t)\) is read "for all t".

 

[HallsofIvy]

The "standard" way to solve "e^{ax}= b" is to take the natural logarithm of both sides: ln(e^{ax})= ax= ln(b) so x= ln(b)/a. Here, however, ln(0) (as well as ln of a negative number) is not defined. That is because e to any power is a positive number. There is NO x that makes e^{-3.5555x}= 0 true.