G
Guest
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can any one help me solve this
4000exp^(-400t)-50000exp^(-1000t)-64000exp^(-1600t)=0
4000exp^(-400t)-50000exp^(-1000t)-64000exp^(-1600t)=0
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4000exp^(-400t)-50000exp^(-1000t)-64000exp^(-1600t)=0
...
\(\displaystyle 50000e^{-1000t} = 4000e^{-400t} - 64000e^{-1600t}\)
Let \(\displaystyle u = e^{-200t}\)
Then:
\(\displaystyle 50000u^5 = 4000u^2 - 64000u^8\)
\(\displaystyle u^2(-64000u^6 + 50000u^3 + 4000) = 0\)
Let \(\displaystyle x = u^3\)
Then:
-\(\displaystyle 1000u^2(64x^2 - 50x - 4) = 0\)
Then set -1000u^2 = 0 and 64x^2 - 50x - 4 = 0
Use quadratic equation from here...
...
\(\displaystyle 50000e^{-1000t} = 4000e^{-400t} - 64000e^{-1600t}\)
Let \(\displaystyle u = e^{-200t}\)
Then:
\(\displaystyle 50000u^5 = 4000u^2 - 64000u^8\)
\(\displaystyle u^2(-64000u^6 + 50000u^3 + 4000) = 0\)
Let \(\displaystyle x = u^3\)
Then:
-\(\displaystyle 1000u^2(64x^2 - 50x - 4) = 0\)
Then set -1000u^2 = 0 and 64x^2 - 50x - 4 = 0
Use quadratic equation from here...
Hello, billybob_bonka!
My approach varies slightly . . .
Multiply by \(\displaystyle e^{^{1600t}}\;\cdots\;2e^{^{1200t}}\,-\,25e^{^{600t}}\,-\,32\;=\;0\)
We have a quadratic: \(\displaystyle \,2(e^{^{600t}})^2\,-\,25e^{^{600t}}\,-\,32\;=\;0\)
Let \(\displaystyle u\,=\,e^{^{600t}}\;\cdots\;2u^2\,-\,25u\,-\,32\;=\;0\)
Quadratic Formula: \(\displaystyle \:u\;=\;\frac{25\,\pm\,\sqrt{25^2\,-\,4(2)(-32)}}{2(2)} \;= \;\frac{25\,\pm\,\sqrt{881}}{4}\)
Then we have: \(\displaystyle \,e^{^{600t}}\:=\;\frac{25\,+\,\sqrt{881}}{4}\;\;\Rightarrow\;\;600t\:=\:\ln\left(\frac{25\,+\,\sqrt{881}}{4}\right)\)
Therefore: \(\displaystyle \:t\;=\;\frac{1}{600}\ln\left(\frac{25\,+\,\sqrt{881}}{4}\right)\;=\;0.004358723...\)
My approach varies slightly . . .
Divide by \(\displaystyle 2,000\;\cdots\;2e^{^{400t}}\,-\,25e^{^{-1000t}}\,-\,32e^{^{-1600t}}\;=\;0\)Can any one help me solve this?
\(\displaystyle 4,000e^{^{-400t}}\,-\,50,000e^{^{-1000t}}\,-\,64,000e^{^{-1600t}}\;=\;0\)
Multiply by \(\displaystyle e^{^{1600t}}\;\cdots\;2e^{^{1200t}}\,-\,25e^{^{600t}}\,-\,32\;=\;0\)
We have a quadratic: \(\displaystyle \,2(e^{^{600t}})^2\,-\,25e^{^{600t}}\,-\,32\;=\;0\)
Let \(\displaystyle u\,=\,e^{^{600t}}\;\cdots\;2u^2\,-\,25u\,-\,32\;=\;0\)
Quadratic Formula: \(\displaystyle \:u\;=\;\frac{25\,\pm\,\sqrt{25^2\,-\,4(2)(-32)}}{2(2)} \;= \;\frac{25\,\pm\,\sqrt{881}}{4}\)
Then we have: \(\displaystyle \,e^{^{600t}}\:=\;\frac{25\,+\,\sqrt{881}}{4}\;\;\Rightarrow\;\;600t\:=\:\ln\left(\frac{25\,+\,\sqrt{881}}{4}\right)\)
Therefore: \(\displaystyle \:t\;=\;\frac{1}{600}\ln\left(\frac{25\,+\,\sqrt{881}}{4}\right)\;=\;0.004358723...\)