\frac{x^2\ln(b) + \log_b(x)}{x\ln(b)\sqrt{x^2 + \log_b(x)^2}} = 0
Added by Chad Y.
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First, we can simplify the expression inside the logarithm by using the product rule of logarithms: logb((x) xln(b)V x) = logb(x) + logb(xln(b)V x) Show more…
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