How can I simplify this further than it is? [8x - {log(base8)x}(8^x)(ln8)] / x(ln8)8^2x
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Originally Posted by theowne How can I simplify this further than it is? [8x - {log(base8)x}(8^x)(ln8)] / x(ln8)8^2x You can't do much. Perhaps canceling an 8 from the top and bottom: $\displaystyle \frac{8x - log_8(x)~8^x~ln(8)}{x~ln(8)~8^{2x}}$ $\displaystyle = \frac{x - log_8(x)~8^{x - 1}~ln(8)}{x~ln(8)~8^{2x - 1}}$ -Dan
What if the first term was changed from 8x to 8^x? I suppose this would change something?
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