Thread: Substituting equation

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

{e}^{2t} - {243}^{t} + 143 = 0

Are you sure the problem is correctly written?
The problem may be
{e}^{2t} - {243}*{e}^{t} + 143 = 0

For any real number, a, a= e^{ln(a)} so, in particular, 243^t= e^{ln(243^t}}= e^{t ln(243)}= (e^t)^{243} so if the equation you give is corect, then you can write it as e^{2t}+ 243^t+ 143= (e^t)^2+ (e^t)^{ln(243)}+ 143= x^2+ x^{ln(243)}+ 143.

So sorry everyone fo this ....

the actual equation was supposed to be

{e}^{2t} - {24e}^{t} + 143 = 0 ...... subing x = {e}^{t}

So very sorry

From the laws of exponents you know that: e^{2t} = (e^t)^2

Thus when you make your sub you get x^2-24x+143=0.

Solve using your favourite method (it will factor but it's probably quicker to use the formula)

Originally Posted by daemonlies

So sorry everyone fo this ....

the actual equation was supposed to be

{e}^{2t} - {24e}^{t} + 143 = 0 ...... subing x = {e}^{t}

So very sorry

I think e^(*pi) got it right, but is that middle term or ?

-Dan

hi dan ... thanks for this ... its supposed to be .... 24e^t

Assuming that it's e^(2t) - 24e^t + 143 = 0, make the substitution x = e^t, giving

x^2 - 24x + 143 = 0.

Now solve the quadratic and then use that solution to solve for t.