Korzystamy ze wzoru:

${\left(c^m\right)}^n=c^{m\cdot n}$

 

Porównujemy potęgi o takich samych podstawach.

$\text{a)}{4}^6>2^{10}$

$4^6={\left(2^2\right)}^6=2^{12}$

$2^{12}>2^{10}$

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$\text{b)}{27}^7>3^{20}$

${27}^7={\left(3^3\right)}^7=3^{21}$

$3^{21}>3^{20}$

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$\text{c)}{25}^4<5^{10}$

${25}^4={\left(5^2\right)}^4=5^8$

$5^8<5^{10}$

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$\text{d)}{125}^3<5^{15}$

${125}^3={\left(5^3\right)}^3=5^9$

$5^9<5^{15}$

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$\text{e)}{8}^9>2^{25}$

$8^9={\left(2^3\right)}^9=2^{27}$

$2^{27}>2^{25}$

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$\text{f)}{100}^7<{10}^{70}$

${100}^7={\left({10}^2\right)}^7={10}^{14}$

${10}^{14}<{10}^{70}$

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$\text{g)}{36}^5=6^{10}$

${36}^5={\left(6^2\right)}^5=6^{10}$

$6^{10}=6^{10}$

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$\text{h)}{10000}^5={10}^{20}$

${10000}^5={\left({10}^4\right)}^5={10}^{20}$

${10}^{20}={10}^{20}$