Question 1022712: what is e^x - 5e^-x - 4 = 0
how to get the appropriate answer hehehe thanks!
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! start with e^x - 5e^-x - 4 = 0
since e^-x is the same as 1/e^x, this becomes e^x - 5/e^x - 4 = 0
multiply both sides of that equation by e^x to get (e^x)^2 - 5 - 4e^x = 0
reorder the terms in descending order of degree to get (e^x)^2 - 4e^x - 5 = 0
this is a quadratic equation where the argument of the function is e^x.
let some other variable name be equal to e^x.
i'll use the variable name of b.
the equation becomes b^2 - 4b - 5 = 0.
now it's easier to see that it is a quadratic equation.
factor this quadratic to get (b-5) * (b+1) = 0
solve for b to get b = 5 or b = -1.
since b is equal to e^x, this becomes e^x = 5 or e^x = -1
solve for x in each of these equations by taking the natural log of both sides of each equation.
you will get ln(e^x) = ln(5) or ln(e^x) = ln(-1).
since you cannot take the log of a negative number and get a real answer, then the only viable solution is ln(e^x) = ln(5).
since ln(e^x) is the same as x * ln(e) and since ln(e) is equal to 1, then you get ln(e^x) is equal to x * 1 which is equal to x.
your equation becomes x = ln(5).
that's your solution.
you can go one step further by finding the natural log of 5 to get x = 1.609437912.
x = ln(5) is the exact solution.
x = 1.609437912 is an approximate solution since it does not appear that ln(5) results in a rational number.
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