Question
Arrange in ascending order:
$2^5, 3^3, 2^3 \times 2,\left(3^3\right)^2, 3^5, 4^0, 2^3 \times 3^1$
$2^5, 3^3, 2^3 \times 2,\left(3^3\right)^2, 3^5, 4^0, 2^3 \times 3^1$
✓
Answer
In ascending order, the numbers are arranged from smallest to largest.
We have, $2^5=2 \times 2 \times 2 \times 2 \times 2=32$
$3^3=3 \times 3 \times 3=27$
$2^3 \times 2=2 \times 2 \times 2 \times 2=16$
$\left(3^3\right)^2=3^3 \times 2\left[\therefore\left(a^m\right)^n=a^{m n}\right]$
$3^6=3 \times 3 \times 3 \times 3 \times 3 \times 3=729$
$3^5=3 \times 3 \times 3 \times 3 \times 3=243$
$4^0=1\left[\therefore a^0=1\right]$
$\text { and } 2^3 \times 3^1$
$=2 \times 2 \times 2 \times 3$
$=24$ Thus, the required ascending order will be.
$4^0<2^3 \times 2<23 \times 31<33<25<35<\left(3^3\right)^2$
We have, $2^5=2 \times 2 \times 2 \times 2 \times 2=32$
$3^3=3 \times 3 \times 3=27$
$2^3 \times 2=2 \times 2 \times 2 \times 2=16$
$\left(3^3\right)^2=3^3 \times 2\left[\therefore\left(a^m\right)^n=a^{m n}\right]$
$3^6=3 \times 3 \times 3 \times 3 \times 3 \times 3=729$
$3^5=3 \times 3 \times 3 \times 3 \times 3=243$
$4^0=1\left[\therefore a^0=1\right]$
$\text { and } 2^3 \times 3^1$
$=2 \times 2 \times 2 \times 3$
$=24$ Thus, the required ascending order will be.
$4^0<2^3 \times 2<23 \times 31<33<25<35<\left(3^3\right)^2$
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