Question 1062567
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solve 25^y+3(5^y)=4
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<pre>
Introduce new variable x = {{{5^y}}}.

Then your equation becomes 

{{{x^2 + 3x - 4}}} = 0.     ( <--- since 25 = {{{5^2}}} )

Factor this quadratic equation:

(x+4)*(x-1) = 0.

Its roots are x= -4 and x= 1.


Hence, we have two equations for y:


1.  {{{5^y}}} = -4.   It HAS NO solutions.


2.  {{{5^y}}} = 1.   It has the solution  y = 0.


<U>Answer</U>. The only solution to the original equation is y = 0.
</pre>

It is the standard method of solving exponential equations.
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.