Question 1094771
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<pre>
Your equation is

{{{5*5^(2x) + 25}}} = {{{125*5^x + 5^x}}},   or, which is the same,

{{{5*5^(2x) - 126*5^x + 25}}} = 0.


Introduce new variable  u = {{{5^x}}}.  Then your equation will take the form

{{{5u^2 - 126u + 25}}} = 0.

Apply the quadratic formula to find the solutions for u:


{{{u[1,2]}}} = {{{(126 +- sqrt(126^2 -4*5*25))/(2*5)}}} = {{{(126 +- 124)/10}}}.

{{{u[1]}}} = {{{(126+124)/10}}} = 25;

{{{u[1]}}} = {{{(126-124)/10}}} = {{{1/5}}}.


From {{{u[1]}}} = 25 = {{{5^x}}} you get the solution for x: x= 2.

From {{{u[2]}}} = {{{1/5}}} = {{{5^x}}} you get another solution for x: x= -1.


<U>Answer</U>.  The original equation has two solutions:  x = 2  and  x = -1.
</pre>

The method of introducing new variable I showed you in this solution, is the standard method for problems like this.



To see more examples of problems solved in this way, see 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.