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.
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.
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.
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.
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:First of all can the -30(5^x) be multiplied? to =-150^x
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.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?
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.
Plug the values of x you solved for back into the original equation to see if they work.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?
Yeah! This is the best method to vertify the answer.Plug the values of x you solved for back into the original equation to see if they work.