# What is Common ?

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?

Tide
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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 dont 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 dont 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.

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.