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Very good, heo83
An addition: Allowing complex numbers, the solution x=-lg(a-1) holds for any a with the only exception a=2, because in this case the step from row 2 to 3 [dividing both numerator and denominator by exp(-x) ] is illegal as exp(-x)=0.
Regards
Good solution thnaks, i also noticed the solution for this case later. I think there is no exact or general solution if we change 2x by 1.2x(or something else) for example, right?
Good solution thnaks, i also noticed the solution for this case later. I think there is no exact or general solution if we change 2x by 1.2x(or something else) for example, right?
the problem is u would get a equation in higher powers to solve..
an alternative solution to the problem (first one).. which can also be applied to the second one asked.. but wth some tedious math involved.. (solving higher power equations)... is...
arunmit168 said:
1-e^-2x/1-e^-x = a
y = e^-x
=> 1-y^2/1-y = a
=> 1 - y^2 = a - ay
=> y^2 -ay + a-1 = 0
solve for this and get two values of y from which only one will suite the solution.. from that takin ln on both sides.. u get teh solution.. here incase if instead of 2 1.2 was there.. then the equation would have been...
y^1.2 = ay - a + 1
=> y^(6/5) = ay - a + 1
=> y^ 6 = (ay - a + 1) ^5
this would go more tedius... instead we can try to solve the initial equation itself.. i am definite its possible to solve that equation!
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