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log5(x+2)-log5 10=log5 100
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Recall, that for any base, \[\log_a b - \log_a c = \log_a\dfrac{b}{c}\]Thus:\[\log_5 \dfrac{x+2}{10} = \log_5 100\]\[\dfrac{x+2}{10}=100\]\[x = 998\]
log5 x+2)/10 = log5 100 (x+2)/10 = 100 x + 2 = 1000 x = 998
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