Question 1144754
.


            It is an advanced level exponential equation.


            Nevertheless,  it is  SOLVABLE  (!),  and I will show you how.



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4^(x+1.5) + 9^(x+0.5) = 10×6^x             (1)


It is equivalent to


2^(2x+3) + 3^(2x+1) = 10*(2^x)*(3^x),   or


8*2^(2x) + 3*3^(2x) = 10*(2^x)*(3^x).      (2) 


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


    8u^2 + 3v^2 = 10uv


Divide both sides by v^2.  You will get


    {{{8*(u/v)^2}}} + 3 - {{{10*(u/v)}}} = 0                (3)


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


    8z^2 - 10z + 3 = 0.


Solve this quadratic equation using the quadratic formula


    {{{z[1,2]}}} = {{{(10 +- sqrt(10^2 - 4*8*3))/(2*8)}}} = {{{(10 +- 2)/16}}}.


The roots are  


    {{{z[1]}}} = {{{(10+2)/16}}} = {{{12/16}}} = {{{3/4}}},   and

    {{{z[2]}}} = {{{(10-2)/16}}} = {{{8/16}}} = {{{1/2}}}.


Thus, we should consider two cases.


(a)  {{{z[1]}}} = {{{3/4}}}.


     It means  {{{3/4}}} = {{{u/v}}} = {{{(2/3)^x}}}.


     Next you can take any logarithm, log base 10, or natural logarithm "ln" to continue


       {{{log(10,(3/4))}}} = {{{x*(log(10,(2/3)))}}},

       {{{x*(log(2) - log(3))}}} = {{{log(3) - log(4)}}}

        x = {{{(log(3) - 2*log(2))/(log(2) - log(3))}}} = {{{(2*log(2) - log(3))/(log(3) - log(2))}}} = 0.709511  (approximately).

The first case is completed.


Similarly, you can consider and complete the second case  {{{z[2]}}} = {{{1/2}}}.


May I leave it to you, in order for you completed it on your own ?
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If you still will have questions, let me know.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Come again to this forum soon to learn something new &nbsp;(!)



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<pre>
As an accurate person, I checked my answer.


I used MS Excel in my computer and calculated the left side and the right side expressions of the original equation at x= 0.709511.


I got the value of  35.65389  on the left side and the value of  35.65389  on the right side.
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