Solve for x given that ½log_{2}9+ log_{2}(5x- 4) = 7

0.5log_{2} 9 + log_{2}(5x − 4) = 7log_{2} 3 + log_{2}(5x − 4) = 7 log_{2} 2log_{2} 3(5x − 4) = log_{2} 2715x = 140 x = 9^{1}/_{3}

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