Solve for t

e^(2t+1)=9e^(1-t)

Question

Solve for t

e^(2t+1)=9e^(1-t)

solomonzelman · Answer

$\large\color{slate}{   e^{2t+1}=9e^{1-t}   }$

anonymous · Answer

i took the natural log of both sides and sort of solved from there but my answers were .2 off so im not sure i did it right

solomonzelman · Answer

yes, you can do something very similar to what you have done as well.

solomonzelman · Answer

For some reason I thought of,

$\large\color{slate}{   e^{2t+1}=\color{blue}{9}e^{1-t}   }$

$\large\color{slate}{   e^{2t+1}=\color{blue}{e^{\ln 9}}e^{1-t}   }$

$\large\color{slate}{   e^{2t+1}=e^{\ln 9+1-t}   }$

anonymous · Answer

hmmm ok. I wouldnt have immediately thought of that but I understand it

anonymous · Answer

Solomon, your final answer for t was around .732, correct?

amoodarya · Answer

sorry , i write another way to solve $$e^{2t+1}=9e^{1-t}\e^{2t}*e=9*e*e^{-t}\e^{2t}=9*e^{-t}\a=e^t\so\a^2=9*\frac{1}{a}\a^3=9$$

solomonzelman · Answer

idk, about the approximation, but:

$\large\color{slate}{   2t+1=\ln(9)+1-t                  }$
$\large\color{slate}{   3t+1=\ln(9)+1                  }$
$\large\color{slate}{   3t=\ln(9)                 }$


$\large\color{slate}{   t=\frac{1}{3}\ln(9)                 }$

anonymous · Answer

Yes thats what I got, Solomon. Thank you.

solomonzelman · Answer

yw