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 ... (Worried)

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- April 28th 2011, 02:08 AMdaemonliesSubstituting 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 ... (Worried) - April 28th 2011, 03:12 AMSpringFan25
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 - April 28th 2011, 03:14 AMsa-ri-ga-ma
**{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 - April 28th 2011, 03:38 AMHallsofIvy
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.

- April 28th 2011, 01:42 PMdaemonlies
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 - April 28th 2011, 01:54 PMe^(i*pi)
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) - April 28th 2011, 02:10 PMtopsquark
- April 28th 2011, 02:45 PMdaemonlies
hi dan ... thanks for this ... its supposed to be .... 24e^t

- April 28th 2011, 11:00 PMProve It
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.