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baherus [9]
3 years ago
7

Help please!!!!!! And explain ur answer.. I will mark as BRAINLIEST​

Mathematics
1 answer:
Temka [501]3 years ago
7 0

Answer:

Nu 4550

Step-by-step explanation:

Commission of 5% only applies to the first 70,000 so

70,000*\frac{5}{100} = 3500

That leaves 15,000 left in sales where the 7% commission applies.

15,000*\frac{7}{100}=1050

Adding these two together we get a total of

3500 +1050=Nu4550

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Write the expression as the sum or difference of logarithms of a, b, and c. Assume all variables represent positive real numbers
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ANSWER

\frac{1}{4}\log _3(a)-\frac{1}{4}\log _3(b)-\frac{5}{4}\log (c^{})

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First we can take the root out of the argument of the logarithm. Remember that roots can be written as fractional exponents:

\sqrt[4]{\frac{a}{bc^5}}=\mleft(\frac{a}{bc^5}\mright)^{1/4}

So applying the power rule of logarithms:

\log _3\sqrt[4]{\frac{a}{bc^5}}=\frac{1}{4}\log _3\frac{a}{bc^5}

Next we can apply the quotient rule of logarithms:

\frac{1}{4}\log _3\frac{a}{bc^5}=\frac{1}{4}\lbrack\log _3(a)-\log _3(bc^5)\rbrack

Then we use the product rule for the last term:

\frac{1}{4}\lbrack\log _3(a)-\log _3(bc^5)\rbrack=\frac{1}{4}\lbrace\log _3(a)-\lbrack\log _3(b)+\log (c^5)\rbrack\rbrace

And the power rule for the exponent of c:

\frac{1}{4}\lbrace\log _3(a)-\lbrack\log _3(b)+\log (c^5)\rbrack\rbrace=\frac{1}{4}\lbrace\log _3(a)-\lbrack\log _3(b)+5\log (c^{})\rbrack\rbrace

What we have to do now is rewrite this to be more clear. Apply the distributive property for the minus sign into the expression with the logarithms of b and c:

\frac{1}{4}\lbrace\log _3(a)-\lbrack\log _3(b)+5\log (c^{})\rbrack\rbrace=\frac{1}{4}\lbrack\log _3(a)-\log _3(b)-5\log (c^{})\rbrack

And then do the same for the 1/4 coefficient:

\frac{1}{4}\lbrack\log _3(a)-\log _3(b)-5\log (c^{})\rbrack=\frac{1}{4}\log _3(a)-\frac{1}{4}\log _3(b)-\frac{5}{4}\log (c^{})

In summary:

\log _3\sqrt[4]{\frac{a}{bc^5}}=\frac{1}{4}\log _3(a)-\frac{1}{4}\log _3(b)-\frac{5}{4}\log (c^{})

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