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Pre-calc, I am so stuck. Please help: Solve: 12^(2x-1) = e^(x+1)
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divide both sides by e^(x + 1) 12^(2x - 1)e^(-1 -x) = 1 (note the sign changes because i moved e^(x + 1) from the denominator to the numerator) write the exponents in 12^(2x - 1)e^(-1 -x) = 1 in terms of a common base 12^(2x - 1)e^(-1 -x) = e^log(12^2x - 1)e^log(e^-1 -x) = e^( (2x - 1)log(12) ) e^( (-1 -x) log(e) ) = e^(-1 -x + (2x - 1) log(12) ) eliminate the exponential from the left hand side: -1 -x + log(12)(2x-1) = 0 clean things up: (2log(12) -1)x = 1 + log(12) x = 1 / (2(log(12) -1) + log(12) / (2log(12) - 1) be mindful of the parentheses in all steps, they are important.
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