Övningar 3.4
Sommarmatte 1
(Skillnad mellan versioner)
Versionen från 16 juli 2007 kl. 11.31 (redigera) KTH.SE:u1zpa8nw (Diskussion | bidrag) ← Gå till föregående ändring |
Nuvarande version (17 juli 2007 kl. 09.43) (redigera) (ogör) KTH.SE:u1zpa8nw (Diskussion | bidrag) |
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(2 mellanliggande versioner visas inte.) | |||
Rad 22: | Rad 22: | ||
<tr align="left"> | <tr align="left"> | ||
<td class="ntext">a) </td> | <td class="ntext">a) </td> | ||
- | <td class="ntext" width="33%">$2^{\scriptstyle x^2-2}=1$</td> | + | <td class="ntext" width="33%">$2^{\scriptstyle x^{\scriptstyle2}-2}=1$</td> |
<td class="ntext">b) </td> | <td class="ntext">b) </td> | ||
<td class="ntext" width="33%">e^{2x}+e^x=4</td> | <td class="ntext" width="33%">e^{2x}+e^x=4</td> | ||
<td class="ntext">c) </td> | <td class="ntext">c) </td> | ||
- | <td class="ntext" width="33%">$3e^{x^2}=2^x$</td> | + | <td class="ntext" width="33%">$3e^{x^{\scriptstyle2}}=2^x$</td> |
</tr> | </tr> | ||
<tr><td height="5px"/></tr> | <tr><td height="5px"/></tr> | ||
Rad 38: | Rad 38: | ||
<tr align="left"> | <tr align="left"> | ||
<td class="ntext">a) </td> | <td class="ntext">a) </td> | ||
- | <td class="ntext" width="50%">$2^{-x^2}=2e^{2x}$</td> | + | <td class="ntext" width="50%">$2^{-x^{\scriptstyle2}}=2e^{2x}$</td> |
<td class="ntext">b) </td> | <td class="ntext">b) </td> | ||
<td class="ntext" width="50%">\ln{(x^2+3x)}=\ln{(3x^2-2x)}</td> | <td class="ntext" width="50%">\ln{(x^2+3x)}=\ln{(3x^2-2x)}</td> | ||
Rad 49: | Rad 49: | ||
</table> | </table> | ||
</div> | </div> | ||
+ | <br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br><br> |
Nuvarande version
Övning 3.4:1
Lös ekvationerna
a) | e^x=13 | b) | 13e^x=2\cdot3^{-x} | c) | 3e^x=7\cdot2^x |
Övning 3.4:2
Lös ekvationerna
a) | 2^{\scriptstyle x^{\scriptstyle2}-2}=1 | b) | e^{2x}+e^x=4 | c) | 3e^{x^{\scriptstyle2}}=2^x |
Övning 3.4:3
Lös ekvationerna
a) | 2^{-x^{\scriptstyle2}}=2e^{2x} | b) | \ln{(x^2+3x)}=\ln{(3x^2-2x)} |
c) | \ln{x}+\ln{(x+4)}=\ln{(2x+3)} | ||