Solution 1.3:6f
From Förberedande kurs i matematik 1
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| - | {{ | + | We can factorize the exponents 40 and 56 as |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>\begin{align} |
| + | 40 &= 4\cdot 10 = 2\cdot 2\cdot 2\cdot 5 = 2^{3}\cdot 5 \\[3pt] | ||
| + | 56 &= 7\cdot 8 = 7\cdot 2\cdot 4 = 7\cdot 2\cdot 2\cdot 2 = 2^{3}\cdot 7 | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | and we then see that they have <math>2^{3} = 8</math> as a common factor. We can take this factor out as an "outer" exponent | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | 3^{40} &= 3^{5\cdot 8} = \bigl(3^{5}\bigr)^{8} = (3\cdot 3\cdot 3\cdot 3\cdot 3)^{8} = 243^{8}\,,\\[3pt] | ||
| + | 2^{56} &= 2^{7\cdot 8} = \bigl(2^{7}\bigr)^{8} = (2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2)^{8} = 128^{8}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | This shows that <math>3^{40} = 243^{8}</math> is bigger than <math>2^{56} = 128^{8}</math>. | ||
Current revision
We can factorize the exponents 40 and 56 as
| \displaystyle \begin{align}
40 &= 4\cdot 10 = 2\cdot 2\cdot 2\cdot 5 = 2^{3}\cdot 5 \\[3pt] 56 &= 7\cdot 8 = 7\cdot 2\cdot 4 = 7\cdot 2\cdot 2\cdot 2 = 2^{3}\cdot 7 \end{align} |
and we then see that they have \displaystyle 2^{3} = 8 as a common factor. We can take this factor out as an "outer" exponent
| \displaystyle \begin{align}
3^{40} &= 3^{5\cdot 8} = \bigl(3^{5}\bigr)^{8} = (3\cdot 3\cdot 3\cdot 3\cdot 3)^{8} = 243^{8}\,,\\[3pt] 2^{56} &= 2^{7\cdot 8} = \bigl(2^{7}\bigr)^{8} = (2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2)^{8} = 128^{8}\,\textrm{.} \end{align} |
This shows that \displaystyle 3^{40} = 243^{8} is bigger than \displaystyle 2^{56} = 128^{8}.
