I'm having trouble substituting this equation ....
{e}^{2t} - {243}^{t} + 143 = 0
substitute x = {e}^{t}
Giving answer to 4 decimal places
Any help would be appreciated ...
are you sure that you typed that correctly?
{e}^{2t} - {243}^{t} + 143 = 0
is hard
{e}^{2t} - {243}e^{t} + 143 = 0
can be solved using the subsititution you suggested.
x^2 - 243x + 143 = 0
... (quadratic formula)...
x = 243.587 (you will need more decimals when you do it)
or
x = -0.58 (ignore this negative solution. Can you see why?)
Finally:
since x = e^t
243.57 = e^t
ln(243.57) = t