anonymous
  • anonymous
Solve the following equations, giving your ans correct to 3 sig. fig. thx u. a) 2e^(2x+1) = e^(x+1) +15e b) 2e^2x = 5e^(x+1) - 2e^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
so how would you go about solving this?
amistre64
  • amistre64
this property of exponents might come in handy:\[\large b^{n+m}=b^n*b^m\]
Zarkon
  • Zarkon
u=e^x for the 2nd

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anonymous
  • anonymous
first one, divide both sides by "e" to get \[2e^{2x}=e^x+15\] then perhaps \[2e^{2x}-e^x-15=0\] a nice quadratic equation, that you can solve by factoring
anonymous
  • anonymous
second one \[2e^{2x} = 5e^{x+1} - 2e^2\] \[2e^{2x}=5e(e^x)-2e^2\] \[2e^{2x}-5e(e^x)+2e^2=0\] quadratic formula for this one

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