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The difference of two logarithms of equal base $b$ is equal to the logarithm of the quotient: $\log_b(x)-\log_b(y)=\log_b\left(\frac{x}{y}\right)$
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$\log \left(\frac{6x-1}{x+4}\right)=\log \left(x\right)$
Learn how to solve problems step by step online. Solve the logarithmic equation log(6*x+-1)-log(x+4)=log(x). The difference of two logarithms of equal base b is equal to the logarithm of the quotient: \log_b(x)-\log_b(y)=\log_b\left(\frac{x}{y}\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. Multiply both sides of the equation by x+4. Solve the product x\left(x+4\right).