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I am having trouble figuring out how to solve this problem.
Find all real solutions to the equation 40-106e^x+101e^2x-41e^3x+6e^4x=0.
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Let us introduce new variable z =
, Then the equation will take a form
p(z) = 6z^4 - 41z^3 +101z^2 - 106z + 40 = 0.
One root is z=1. You can check it making direct calculation.
So, the polynomial has the divisor (z-1).
The quotient
is q(z) =
(long division).
The root of the cubic polynomial is z=2. You can check it making direct calculation.
So, the polynomial has the divisor (z-2).
The quotient
is r(z) =
(long division).
r(z) has the roots 2.5 and
(quadratic formula).
So, the polynomial p(z) has the roots 1, 2, 2.5 and
.
Since z =
, the original polynomial has the roots
x = ln(1) = 0, x = ln(2), x = ln(2.5) and x = ln((3/4)).
Solved.