3.4 Övningar

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{{Mall:Ej vald flik|[[3.4 Logaritmekvationer|Teori]]}}
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{{Ej vald flik|[[3.4 Logaritmekvationer|Theory]]}}
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{{Mall:Vald flik|[[3.4 Övningar|Övningar]]}}
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{{Vald flik|[[3.4 Övningar|Exercises]]}}
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===Övning 3.4:1===
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===Exercise 3.4:1===
<div class="ovning">
<div class="ovning">
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Best&auml;m <math>\,x\,</math> om
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Solve the equation
{| width="100%" cellspacing="10px"
{| width="100%" cellspacing="10px"
|a)
|a)
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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:Svar|Svar 3.4:1|Lösning a|Lösning 3.4:1a|Lösning b|Lösning 3.4:1b|Lösning c|Lösning 3.4:1c}}
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</div>{{#NAVCONTENT:Answer|Svar 3.4:1|Solution a|Lösning 3.4:1a|Solution b|Lösning 3.4:1b|Solution c|Lösning 3.4:1c}}
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===Exercise 3.4:2===
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<div class="ovning">
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Solve the equation
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{| width="100%" cellspacing="10px"
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|a)
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|width="33%" | <math>2^{\scriptstyle x^2-2}=1</math>
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|b)
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|width="33%" | <math>e^{2x}+e^x=4</math>
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|c)
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|width="33%" | <math>3e^{x^2}=2^x</math>
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|}
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</div>{{#NAVCONTENT:Answer|Svar 3.4:2|Solution a|Lösning 3.4:2a|Solution b|Lösning 3.4:2b|Solution c|Lösning 3.4:2c}}
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===Exercise 3.4:3===
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<div class="ovning">
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Solve the equation
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{| width="100%" cellspacing="10px"
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|a)
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|width="50%" | <math>2^{-x^2}=2e^{2x}</math>
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|b)
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|width="50%" | <math>\ln{(x^2+3x)}=\ln{(3x^2-2x)}</math>
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|-
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|c)
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|width="50%" | <math>\ln{x}+\ln{(x+4)}=\ln{(2x+3)}</math>
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|}
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</div>{{#NAVCONTENT:Answer|Svar 3.4:3|Solution a|Lösning 3.4:3a|Solution b|Lösning 3.4:3b|Solution c|Lösning 3.4:3c}}

Aktuelle Version

       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
Answer
Svar 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

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
Answer
Svar 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

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)}
Answer
Svar 3.4:3
Solution a
Lösning 3.4:3a
Solution b
Lösning 3.4:3b
Solution c
Lösning 3.4:3c
Von „http://wiki.sommarmatte.se/wikis/2008/bridgecourse1-TU-Berlin/index.php/3.4_%C3%96vningar“