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9^x = 3^(1-2x) What is x? Explain to me how to do these ridiculous problems with the exponents being all subtracted.
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trick is to write with the same base, then compare exponents
you have \[9^x\] and \[9=3^2\] so this is \[(3^2)^x=3^{2x}\]
so the left hand side is \[3^{2x}\] right hand is \[3^{1-2x}\] so \[3^{2x}=3^{1-2x}\] ,making \[2x=1-2x\] solve for x
point is not the exponents being subtracted, point is making them equal to see what you get for x. you get \[4x=1\] \[x=\frac{1}{4}\]
You're the best, thank you! Makes much more sense now!
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thx and yw
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