3.3 Übungen
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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|width="50%" | <math>\displaystyle \frac{1}{10^x}=0\textrm{.}000\,1</math> | |width="50%" | <math>\displaystyle \frac{1}{10^x}=0\textrm{.}000\,1</math> | ||
|} | |} | ||
| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Antwort|Antwort 3.3:1|Solution a|Solution 3.3:1a|Solution b|Solution 3.3:1b|Solution c|Solution 3.3:1c|Solution d|Solution 3.3:1d}} |
===Übung 3.3:2=== | ===Übung 3.3:2=== | ||
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|width="25%" | <math>\lg{\displaystyle \frac{1}{10^2}}</math> | |width="25%" | <math>\lg{\displaystyle \frac{1}{10^2}}</math> | ||
|} | |} | ||
| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Antwort|Antwort 3.3:2|Solution a|Solution 3.3:2a|Solution b|Solution 3.3:2b|Solution c|Solution 3.3:2c|Solution d|Solution 3.3:2d|Solution e|Solution 3.3:2e|Solution f|Solution 3.3:2f|Solution g|Solution 3.3:2g|Solution h|Solution 3.3:2h}} |
===Übung 3.3:3=== | ===Übung 3.3:3=== | ||
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|width="33%" | <math>\log_a{\bigl(a^2\sqrt{a}\,\bigr)}</math> | |width="33%" | <math>\log_a{\bigl(a^2\sqrt{a}\,\bigr)}</math> | ||
|} | |} | ||
| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Antwort|Antwort 3.3:3|Solution a|Solution 3.3:3a|Solution b|Solution 3.3:3b|Solution c|Solution 3.3:3c|Solution d|Solution 3.3:3d|Solution e|Solution 3.3:3e|Solution f|Solution 3.3:3f|Solution g|Solution 3.3:3g|Solution h|Solution 3.3:3h}} |
===Übung 3.3:4=== | ===Übung 3.3:4=== | ||
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|width="33%" | <math>\lg{27^{1/3}}+\displaystyle \frac{\lg{3}}{2}+\lg{\displaystyle \frac{1}{9}}</math> | |width="33%" | <math>\lg{27^{1/3}}+\displaystyle \frac{\lg{3}}{2}+\lg{\displaystyle \frac{1}{9}}</math> | ||
|} | |} | ||
| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Antwort|Antwort 3.3:4|Solution a|Solution 3.3:4a|Solution b|Solution 3.3:4b|Solution c|Solution 3.3:4c}} |
===Übung 3.3:5=== | ===Übung 3.3:5=== | ||
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|width="33%" | <math>\left(e^{\ln{e}}\right)^2</math> | |width="33%" | <math>\left(e^{\ln{e}}\right)^2</math> | ||
|} | |} | ||
| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Antwort|Antwort 3.3:5|Solution a|Solution 3.3:5a|Solution b|Solution 3.3:5b|Solution c|Solution 3.3:5c|Solution d|Solution 3.3:5d|Solution e|Solution 3.3:5e|Solution f|Solution 3.3:5f}} |
===Übung 3.3:6=== | ===Übung 3.3:6=== | ||
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||{{LOGCALCULATOR}} | ||{{LOGCALCULATOR}} | ||
|} | |} | ||
| - | </div>{{#NAVCONTENT: | + | </div>{{#NAVCONTENT:Antwort|Antwort 3.3:6|Solution a|Solution 3.3:6a|Solution b|Solution 3.3:6b|Solution c|Solution 3.3:6c}} |
Version vom 09:25, 22. Okt. 2008
Übung 3.3:1
What is \displaystyle \,x\, if
| a) | \displaystyle 10^x=1\,000 | b) | \displaystyle 10^x=0\textrm{.}1 |
| c) | \displaystyle \displaystyle \frac{1}{10^x}=100 | d) | \displaystyle \displaystyle \frac{1}{10^x}=0\textrm{.}000\,1 |
Antwort
Laden...
Solution a
Solution b
Solution c
Solution d
Übung 3.3:2
Calculate
| a) | \displaystyle \lg{ 0\textrm{.}1} | b) | \displaystyle \lg{ 10\,000} | c) | \displaystyle \lg {0\textrm{.}001} | d) | \displaystyle \lg {1} |
| e) | \displaystyle 10^{\lg{2}} | f) | \displaystyle \lg{10^3} | g) | \displaystyle 10^{-\lg{0\textrm{.}1}} | h) | \displaystyle \lg{\displaystyle \frac{1}{10^2}} |
Antwort
Solution a
Solution b
Solution c
Solution d
Solution e
Solution f
Solution g
Solution h
Übung 3.3:3
Calculate
| a) | \displaystyle \log_2{8} | b) | \displaystyle \log_9{\displaystyle \frac{1}{3}} | c) | \displaystyle \log_2{0\textrm{.}125} |
| d) | \displaystyle \log_3{\left(9\cdot3^{1/3}\right)} | e) | \displaystyle 2^{\log_{\scriptstyle2}{4}} | f) | \displaystyle \log_2{4}+\log_2{\displaystyle \frac{1}{16}} |
| g) | \displaystyle \log_3{12}-\log_3{4} | h) | \displaystyle \log_a{\bigl(a^2\sqrt{a}\,\bigr)} |
Antwort
Solution a
Solution b
Solution c
Solution d
Solution e
Solution f
Solution g
Solution h
Übung 3.3:4
Simplify
| a) | \displaystyle \lg{50}-\lg{5} | b) | \displaystyle \lg{23}+\lg{\displaystyle \frac{1}{23}} | c) | \displaystyle \lg{27^{1/3}}+\displaystyle \frac{\lg{3}}{2}+\lg{\displaystyle \frac{1}{9}} |
Antwort
Solution a
Solution b
Solution c
Übung 3.3:5
Simplify
| a) | \displaystyle \ln{e^3}+\ln{e^2} | b) | \displaystyle \ln{8}-\ln{4}-\ln{2} | c) | \displaystyle (\ln{1})\cdot e^2 |
| d) | \displaystyle \ln{e}-1 | e) | \displaystyle \ln{\displaystyle \frac{1}{e^2}} | f) | \displaystyle \left(e^{\ln{e}}\right)^2 |
Antwort
Solution a
Solution b
Solution c
Solution d
Solution e
Solution f
Übung 3.3:6
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Use the calculator on the right to calculate the following to three decimal places. (The button LN signifies the natural logarithm with base e):
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Antwort
Solution a
Solution b
Solution c
