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

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The equation 25^x - 30(5^x) + 125 = 0 can be simplified by substituting A = 5^x, transforming it into A^2 - 30A + 125 = 0. Factoring this gives (A - 5)(A - 25) = 0, leading to solutions A = 5 or A = 25. Consequently, the values of x are 1 and 2, indicating that there can be two valid solutions. To confirm accuracy, it is recommended to substitute these values back into the original equation.
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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 25 = 5^2
 
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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