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Alja [10]
3 years ago
6

Step by step 7 ,9, and 11 please. EVEN IF U CANT DO ALL, u can help with 1 or 2. Ill mark brainly.

Mathematics
1 answer:
o-na [289]3 years ago
8 0

Answer:

7) 4 \ log_3(x) - 4 \ log_3(y)

9) 5log_4(7) - 5log_4(12)

11) 5log_5 \ (x) - log_5 \ (y)

Step-by-step explanation:

log_3 (\frac{x}{y})^{4}

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Use Logarithm of a Quotient which states

log_b \frac{M}{N}  = log_b M-log_bN

And also use Logarithm of a Power which states

log_b\ M^{n} = n\log_bM

----------------------------------------------------------------------------------------------------

So using these two properties,

7. 4 \ log_3(x) - 4 \ log_3(y)

----------------------------------------------------------------------------------------------------

----------------------------------------------------------------------------------------------------

For #9, use the same logarithm propertied

log_4(\frac{7}{12})^5 = 5log_4(7) - 5log_4(12)

----------------------------------------------------------------------------------------------------

----------------------------------------------------------------------------------------------------

#11 is also the same concept

log_5\ \frac{x^5}{y} = 5log_5 \ (x) - log_5 \ (y)

It is not  - 5 log5(y) since only x is to the power of 5 not y

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3 years ago
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IceJOKER [234]
\bf 343^{\frac{2}{3}}+36^{\frac{1}{2}}-256^{\frac{3}{4}}\qquad \begin{cases}
343=7\cdot 7\cdot 7\\
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36=6\cdot 6\\
\qquad 6^2\\
256=4\cdot 4\cdot 4\cdot 4\\
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\end{cases}\\\\\\ (7^3)^{\frac{2}{3}}+(6^2)^{\frac{1}{2}}-(4^4)^{\frac{3}{4}}
\\\\\\
\sqrt[3]{(7^3)^2}+\sqrt[2]{(6^2)^1}-\sqrt[4]{(4^4)^3}\implies \sqrt[3]{(7^2)^3}+\sqrt[2]{(6^1)^2}-\sqrt[4]{(4^3)^4}
\\\\\\
7^2+6-4^3\implies 49+6-64\implies -9


to see what you can take out of the radical, you can always do a quick "prime factoring" of the values, that way you can break it in factors to see who is what.
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