Skillnad mellan versioner av "1.5 Lösning 4c"

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m (Created page with "<math> {9\,^{z+1} \cdot 81\,^{3\,z/4} \over 27\,^{5\,z/3}} = {(3^2)^{z+1} \cdot (3^4)^{3\,z/4} \over (3^3)^{5\,z/3}} = {3^{2\cdot(z+1)} \cdot 3^{4\cdot(3\,z/4)} \over 3^{3\cdot(5...")
 
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<math> {9\,^{z+1} \cdot 81\,^{3\,z/4} \over 27\,^{5\,z/3}} = {(3^2)^{z+1} \cdot (3^4)^{3\,z/4} \over (3^3)^{5\,z/3}} = {3^{2\cdot(z+1)} \cdot 3^{4\cdot(3\,z/4)} \over 3^{3\cdot(5\,z/3)}} = </math>
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::<math> 5,07 \cdot 10\,^6 \, = \, 5,07 \cdot 1\,000\,000 \, = \, 5\,070\,000 </math>
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<math> = {3^{2\,z+2} \cdot 3^{3\,z} \over 3^{5\,z}} = {3^{2\,z+2} \cdot 3^{3\,z} \over 3^{5\,z}} = {3^{2\,z+2+3\,z} \over 3^{5\,z}} = {3^{5\,z+2} \over 3^{5\,z}} = {3^{5\,z+2-5\,z}} = 3^2 = 9 </math>
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Nuvarande version från 7 juli 2015 kl. 15.30

\[ 5,07 \cdot 10\,^6 \, = \, 5,07 \cdot 1\,000\,000 \, = \, 5\,070\,000 \]
Hämtad från "https://matte1b.mathonline.se/index.php?title=1.5_Lösning_4c&oldid=26663"