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solve for x: 9^x = e^(x+1)
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\[9^{x}=e ^{(x+1)}\]\[3^{2(x)}=e ^{(x+1)}\]\[2x \log(3)=(x+1)\log\]\[2xlog(3)=x \log +\log\]\[2x \log(3)-x \log =\log\]\[x(2\log(3)-1) = 1\]\[x=\frac{1}{2\log(3)-1}\]=0.835265
check some one ansewr frist I'm not sure
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