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Multiply both sides of the equation by $\log \left(3x-4\right)$
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$\log \left(16-x^2\right)=2\log \left(3x-4\right)$
Learn how to solve integral calculus problems step by step online. Solve the logarithmic equation log(16+-1*x^2)/log(3*x+-4)=2. Multiply both sides of the equation by \log \left(3x-4\right). Apply the formula: a\log_{b}\left(x\right)=\log_{b}\left(x^a\right). For two logarithms of the same base to be equal, their arguments must be equal. In other words, if \log(a)=\log(b) then a must equal b. Expand \left(3x-4\right)^2.