Value of x: 25^x-30(5^x)+125=0 Common Base?

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Homework Help Overview

The discussion revolves around solving the equation 25^x - 30(5^x) + 125 = 0, focusing on identifying a common base for the terms involved. Participants explore the implications of substituting variables and simplifying the equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants consider the possibility of substitution and simplification, with one suggesting letting A = 5^x. There is also a discussion about the validity of multiplying terms and the implications of the equation's form.

Discussion Status

The conversation has progressed to exploring potential solutions after factoring the equation. Some participants express uncertainty about the correctness of their steps and the possibility of multiple values for x, while others suggest verifying the solutions by substituting back into the original equation.

Contextual Notes

There is a mention of confusion regarding the multiplication of terms and the interpretation of the equation's structure, indicating a need for clarity on the algebraic manipulations involved.

aisha
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Find the exact value of x: 25^x-30(5^x)+125=0 what is the common base? I thought it was 5 but not for 30.
 
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Have a look at the form of the equation... can you do a subsitution?
 
HINT: Observe that [itex]25 = 5^2[/itex]
 
Also, simplify the equation first.
 
First of all can the -30(5^x) be multiplied? to =-150^x
Next I let A=5^x and then my equation became A^2-30+125=0 so A^2+95 I don't think I did this right and if I did then what do I do next A=square root of -95?
 
aisha said:
First of all can the -30(5^x) be multiplied? to =-150^x
Nope. Observe that 2*3^4 = 2*(3^4) = 2*3*3*3*3. On the other hand, (2*3)^4 = 2*2*2*2*3*3*3*3 which is definitely not equivalent.
aisha said:
Next I let A=5^x and then my equation became A^2-30+125=0 so A^2+95 I don't think I did this right and if I did then what do I do next A=square root of -95?
Note that your equation can be written (5^x)^2 - 30(5^x) + 125 = 0, so if A = 5^x, the equation becomes A^2 - 30A + 125 = 0.
 
hypermorphism said:
Nope. Observe that 2*3^4 = 2*(3^4) = 2*3*3*3*3. On the other hand, (2*3)^4 = 2*2*2*2*3*3*3*3 which is definitely not equivalent.

Note that your equation can be written (5^x)^2 - 30(5^x) + 125 = 0, so if A = 5^x, the equation becomes A^2 - 30A + 125 = 0.

Ok I factored that and got (A-5) (A-25) A=5 or A=25 sooo 5^x=5 or 5^x=25
x=1 or x=2 are my solutions correct? Can there be two values for x?
 
aisha said:
Ok I factored that and got (A-5) (A-25) A=5 or A=25 sooo 5^x=5 or 5^x=25
x=1 or x=2 are my solutions correct? Can there be two values for x?
Plug the values of x you solved for back into the original equation to see if they work. :smile:
 
Plug the values of x you solved for back into the original equation to see if they work.
Yeah! This is the best method to vertify the answer.
 

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