a)

${\left(2^4\right)}^3=2^{4\cdot 3}=2^{12}$

${\left(2^9\right)}^2=2^{9\cdot 2}=2^{18}$

$2^{12}<2^{18},$ zatem:

${\left(2^4\right)}^3<{\left(2^9\right)}^2$

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b)

${\left({\left(-0,23\right)}^2\right)}^4={\left(-0,23\right)}^{2\cdot 4}={\left(-0,23\right)}^8=0,23^8$

$0,23^8<0,23^7,$ zatem:

${\left({\left(-0,23\right)}^2\right)}^4<0,23^7$

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c)

${\left({\left(1\frac{1}{5}\right)}^0\right)}^3=1^3=1$

${\left({\left(1\frac{1}{5}\right)}^2\right)}^0=1$

Zatem:

${\left({\left(1\frac{1}{5}\right)}^0\right)}^3={\left({\left(1\frac{1}{5}\right)}^2\right)}^0$

Komentarz: Dowolna liczba podniesiona do potęgi 0 zawsze da w wyniku 1.

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d)

${\left({\left(12^5\right)}^6\right)}^2=12^{5\cdot 6\cdot 2}=12^{60}$

${\left({\left(12^2\right)}^5\right)}^6=12^{2\cdot 5\cdot 6}=12^{60}$

Zatem:

${\left({\left(12^5\right)}^6\right)}^2={\left({\left(12^2\right)}^5\right)}^6$