Question 1143609
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<pre>

    {{{e^(2x+1) + 9e^x - 11}}} = 0      (1)


    {{{e*e^(2x) + 9*e^x - 11}}} = 0      (2)


Introduce new variable t = e^x.

Then the equation (2)  takes the form


     {{{e*t^2 + 9t - 11}}} = 0.       (3)


Apply the quadratic formula


     t = {{{(-9 +- sqrt(9^2 - 4*e*(-11)))/(2*e)}}} = {{{(-9 +- sqrt(81 + 44e))/(2e)}}}.


So, the equation (3) has two roots


      {{{t[1]}}} = {{{(-9 + sqrt(81 + 44e))/(2e)}}}  and  {{{t[2]}}} = {{{(-9 - sqrt(81 + 44e))/(2e)}}}.


Both the roots  {{{t[1]}}}  and  {{{t[2]}}} are real numbers.


Since  t = {{{e^x}}}, we can use ONLY POSITIVE root {{{t[1]}}} to find x:


    x = {{{ln((-9 + sqrt(81 + 44e))/(2e))}}}.      <U>ANSWER</U>


It is the ONLY SOLUTION to the original equation (1).
</pre>

Solved.


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Introducing new variable is the STANDARD method for solving such equations.


To see other similar solved problems, look into the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/logarithm/How-to-solve-exponential-equations.lesson>Solving exponential equations</A>

in this site.