3.4 Exercises
From Förberedande kurs i matematik 1
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|width="33%" | <math>3e^x=7\cdot2^x</math> | |width="33%" | <math>3e^x=7\cdot2^x</math> | ||
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| - | </div>{{#NAVCONTENT:Answer| | + | </div>{{#NAVCONTENT:Answer|Answer 3.4:1|Solution a|Lösning 3.4:1a|Solution b|Lösning 3.4:1b|Solution c|Lösning 3.4:1c}} |
===Exercise 3.4:2=== | ===Exercise 3.4:2=== | ||
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|width="33%" | <math>3e^{x^2}=2^x</math> | |width="33%" | <math>3e^{x^2}=2^x</math> | ||
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| - | </div>{{#NAVCONTENT:Answer| | + | </div>{{#NAVCONTENT:Answer|Answer 3.4:2|Solution a|Lösning 3.4:2a|Solution b|Lösning 3.4:2b|Solution c|Lösning 3.4:2c}} |
===Exercise 3.4:3=== | ===Exercise 3.4:3=== | ||
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|width="50%" | <math>\ln{x}+\ln{(x+4)}=\ln{(2x+3)}</math> | |width="50%" | <math>\ln{x}+\ln{(x+4)}=\ln{(2x+3)}</math> | ||
|} | |} | ||
| - | </div>{{#NAVCONTENT:Answer| | + | </div>{{#NAVCONTENT:Answer|Answer 3.4:3|Solution a|Lösning 3.4:3a|Solution b|Lösning 3.4:3b|Solution c|Lösning 3.4:3c}} |
Revision as of 07:09, 9 September 2008
| Theory | Exercises |
Exercise 3.4:1
Solve the equation
| a) | \displaystyle e^x=13 | b) | \displaystyle 13e^x=2\cdot3^{-x} | c) | \displaystyle 3e^x=7\cdot2^x |
Loading...Exercise 3.4:2
Solve the equation
| a) | \displaystyle 2^{\scriptstyle x^2-2}=1 | b) | \displaystyle e^{2x}+e^x=4 | c) | \displaystyle 3e^{x^2}=2^x |
Exercise 3.4:3
Solve the equation
| a) | \displaystyle 2^{-x^2}=2e^{2x} | b) | \displaystyle \ln{(x^2+3x)}=\ln{(3x^2-2x)} |
| c) | \displaystyle \ln{x}+\ln{(x+4)}=\ln{(2x+3)} |
