Question 1150869
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<pre>

The given exponential equation


    {{{4^x}}} + {{{6^x}}} = {{{9^x}}}             (1)


is equivalent to


    {{{2^(2x)}}} + {{{6^x)}}} = {{{3^(2x)}}},   or


    {{{2^(2x)}}} + {{{2^x*3^x}}} = {{{3^(2x)}}}.          (2) 


Introduce new variables  u = 2^x,  v = 3^x.  Then equation (2) takes the form

    u^2 + u*v = {{{v^2}}}.


Divide both sides by u^2.  You will get

    {{{(v/u)^2}}} - {{{v/u}}} - 1 = 0                (3)


Let z = {{{v/u}}}.   Then equation (3) takes the form

    z^2 - z - 1 = 0.


Solve this quadratic equation using the quadratic formula


    {{{z[1,2]}}} = {{{(1 +- sqrt((-1)^2 + 4))/2}}} = {{{(1 +- sqrt(5))/2}}}.


The roots are  

    {{{z[1]}}} = {{{(1 + sqrt(5))/2}}},   and

    {{{z[2]}}} = {{{(1 - sqrt(5))/2}}}.


Thus, we should consider two cases.


(a)  {{{z[1]}}} = {{{(1+sqrt(5))/2)}}}.


     It means  {{{(1+sqrt(5))/2)}}} = {{{v/u}}} = {{{(3/2)^x}}}.


     Next, take any logarithm, log base 10, or natural logarithm "ln"  from both sides to continue


       {{{log(10, ((1+sqrt(5))/2))}}} = {{{x*(log(10,(3/2)))}}},

        x = {{{log (((sqrt(5)+1)/2))}}} / {{{log((3/2))}}} =  = 1.1868  (approximately).


    Thus this case is completed.



(b)  {{{z[2]}}} = - {{{(sqrt(5)+1)/2)}}}


     It means  - {{{(sqrt(5)+1)/2)}}} = {{{v/u}}} = {{{(3/2)^x}}}.

     
     The left side is negative, while the right side is positive.


     So, this case has no solutions.
     

<U>ANSWER</U>.  The original equation has only one root  x = {{{log (((sqrt(5)+1)/2))}}} / {{{log((3/2))}}} =  = 1.1868  (approximately).
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On solving exponential equations see also 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>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/logarithm/OVERVIEW-of-lessons-on-solving-exponential-equations.lesson>OVERVIEW of lessons on solving exponential equations</A>

in this site.