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Solve : 3^4x-7= 4^2x+3
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this a repeat?
Yep it is...let me find it
start with \[(4x-7)\ln(3)=(2x+3)\ln(4)\] then algebra
\[4\ln(3)x-7\ln(3)=2\ln(4)x-3\ln(4)\] \[4\ln(3)x-2\ln(4)x=7\ln(3)-3\ln(4)\] \[(4\ln(x)-2\ln(4))x = 7\ln(3)-3\ln(4)\] \[x=\frac{7\ln(3)-3\ln(4)}{4\ln(x)-2\ln(4)\]
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\[x=\frac{7\ln(3)-3\ln(4)}{4\ln(x)-2\ln(4)}\]
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