Question 1059243
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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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<pre>
Let us introduce new variable z = {{{e^x}}}, 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 {{{p(z)/(z-1)}}} is q(z) = {{{6z^3 - 35z^2 + 66z - 40}}} (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 {{{q(z)/(z-2)}}} is r(z) = {{{6z^2 - 23x + 20}}} (long division).


r(z) has the roots 2.5 and {{{4/3}}} (quadratic formula).


So, the polynomial p(z) has the roots  1, 2, 2.5 and {{{4/3}}}.


Since z = {{{e^x)}}}, the original polynomial has the roots 


x = ln(1) = 0, x = ln(2), x = ln(2.5) and x = ln((3/4)).
</pre>

Solved.