Justify conclusion on why 2 ounces of ketchup is ok to use when eating potatoe wedges

1 answer:

4 0

Well you haven't really given us a question that requires a "quantifiable" answer. 2 ounces is roughly equivalent to 60 grams. 2 ounces of ketchup wouldn't be an awful lot of ketchup. I'm sure that you'd remain healthy despite eating 2 ounces of ketchup along with your potato wedges.

According to the American Heart Association, women should only have about 25 grams of sugar per day. In 100g of ketchup there would be about 22g grams of sugar. If you were to eat roughly 60g of ketchup, you'd only be digesting about 13.2 grams of sugar. Now, potato wedges could also contain sugar, but you haven't stated how many grams or kilograms of potato wedges you'd be eating, so I wouldn't be able to tell you whether your meal contains too much sugar.

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$\log_x(y) = \dfrac{\ln(y)}{\ln(x)}$

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$xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}$

Use the product-to-sum identity,

$\log_x(yz) = \log_x(y) + \log_x(z)$

to write

$xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}$

Redistribute the factors on the left side as

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and simplify to

$xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)$

Now expand the right side:

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}$

Simplify and rewrite using the logarithm properties mentioned earlier.

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1$