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\[3^x+3-9^{2x}-1=0?\]

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

second one

Ok did you add to both sides \[9^{2x+1}\]

\[3^{x+2}=9^{2x+1}\]

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

\[x+2=2(2x-1)\]

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