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Log8(_)-log8 7 = log8 5/7 fill in the blank (_)
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Log8(_)-log8 7 = log8 5/7

fill in the blank (_)

User Mac Adada
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\begin{array}{llll} \textit{Logarithm of rationals} \\\\ \log_a\left( (x)/(y)\right)\implies \log_a(x)-\log_a(y) \end{array} \\\\[-0.35em] ~\dotfill\\\\ \log_8(x)-\log_8(7)=\log_8\left( \cfrac{5}{7} \right)\implies \log_8(\stackrel{x }{5})-\log_8(7)=\log_8\left( \cfrac{5}{7} \right)

User Michael Isvy
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