25^x-4=5^3x+1

Question

i need it for my recovery packet

Answer 1

I believe it is probably the second option. If you play with the equation you can get the bases to be the same, at which point you can set the exponents equal to each other and solve for x.  Since that answer comes out to a whole number I think that is probably your equation.

Answer 2

25^(x-4)=5^(3x+1)

(5²)^(x-4)=5^(3x+1)

5^(2x-8)=5^(3x+1)

2x-8=3x+1

2x-3x=8+1

-x=9

x=-9

check...

25^(-9-4)=5^(3(-9)+1)

25^(-13)=5^(-27+1)

25^(-13)=5^(-26)

(5²⁾^(-13)=5^(-26)

5^(-26)=5^(-26)

Answer 3

I am going to assume the second interpretation of the two that Philip mentions.

 

I think the easiest way to solve is to rewrite 25 as 5^2 and then observe (due to multiplication of expos) that the problem can be re-written as 5^(2x-2)=5^(3x+1).  we can apply the log based 5 to both sides, thereby inverting the expo function and are left solving a one-variable linear equation:

 

2x-2=3x+1

 

From here, use algebra 1 techniques.