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3^X+3 - 9^2X-1=0 what are the steps to finding the solution?

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Or! Do you mean \[3^{x+3}-9^{2x-1}=0\]
second one

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Other answers:

Ok did you add to both sides \[9^{2x+1}\]
oops - not + oh well
Do you know how to answer the following What number satisfies the equation \[9=3^{?}\]
Ok great. So we have \[3^{x+2}=(3^{2})^{(2x-1)}\] since 3^2=9 :)
So we have the same base on both sides All you need to do is set the exponents equal and solve for x
Let me know if you have further questions :)
firstly its x+3 but where did you get the 2 in "2"(2x-1)
\[9=3^2 \] \[3^{x+3}=9^{2x-1}\] \[3^{x+2}=(3^2)^{(2x-1)}\]
how does 3^2(^2x-1) = 2(2x-1)
No....we have x+2=2(2x-1) not what you wrote

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