Logaritme- og eksponentiallikninger

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Question statement

Løs likningene. Avrund svaret til 2 desimaler.

Give any introductory information the student needs.

			Name	Type	Generated Value

a1	integer	2
b1	integer	-2
c1	integer	-1
a2	integer	1
b2	integer	10
c2	integer	-7
x	list	[ -10.656854, 0.656854 ]
sol	list	[ 0.00, 1.93 ]
disc	number	128
se1	number	-0.6931471806
e1	integer	2
disc2	number	9
x2	list	[ 2.000000, 5.000000 ]
sol2	list	[ 0.69, 1.61 ]
logn	string	(ln(x))

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xxxxxxxxxx
 
random(-4..4 except 0) 

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Generated value: `integer`

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Advice
"Part a)" - prompt
"Part a)" → "Unnamed gap" - Minimum accepted value
"Part a)" → "Unnamed gap" - Maximum accepted value

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Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

$\simplify{{a1} ln(x)+{b1}={c1}}$

$x=$ Unnamed gap

Advice

Løsningsformelen for en logaritmelikning på grunnform ser slik ut:

$\begin{align}\ln(x)&=a\\ &\Downarrow \\x&=e^{a}\end{align}$

Vi snur på likningen og regner ut:

$\begin{align} \simplify{{a1} ln(x)+{b1}}&=\var{c1}\\ \simplify{{a1}ln(x)} &= \simplify[!collectNumbers]{{c1}-{b1}}\\ \simplify{{a1}ln(x)}&= \simplify{{c1}-{b1}}\\ \ln(x) &= \simplify{{{c1}-{b1}}/{{a1}}}\\ x &= e^{\simplify{{{c1}-{b1}}/{{a1}}}}\\ \end{align}$

$x=$ {precround({exp(({c1}-{b1})/{a1})},2)}

Likningen $\simplify{{a2}(ln(x))^{2}+{b2}ln(x)+{c2}=0}$ er en andregradslikning med $\ln(x)$ som ukjent. Vi bruker da først abc-formelen for å finne $\ln(x)$ :

$a=\var{a2},\;b=\var{b2},\;c=\var{c2}$

$\ln(x)=\dfrac{\simplify{-{b2}}\pm\sqrt{\simplify[!collectNumbers,timesDot]{{b2}^{2}-4*{a2}*{c2}}}}{2\cdot\var{a2}}=\dfrac{\simplify{-{b2}}\pm\sqrt{\simplify{{b2}^2-4*{a2}*{c2}}}}{\simplify{2*{a2}}}$ .

Vi to løsninger for $\ln(x)$ : $\ln(x)= \var{x[0]}\text{ og } \ln(x)=\var{x[1]}$ .

Vi løser for $x$ :

$\ln(x)= \var{x[0]}\Rightarrow x=e^{\var{x[0]}}=$ {precround(e^(x[0]),6)} $\approx\underline{\underline{\var{sol[0]}}}$

$\ln(x)= \var{x[1]}\Rightarrow x=e^{\var{x[1]}}=$ {precround(e^(x[1]),6)} $\approx\underline{\underline{\var{sol[1]}}}$

Løsningsformelen for eksponentiallikning på grunnform ser slik ut:

$\begin{align}e^{x}&=a\\ &\Downarrow \\x&=\ln(a)\end{align}$

Vi snur på likningen og regner ut:

$\begin{align} e^{\simplify{{c1}x}}-\var{e1}&=0\\ e^{\simplify{{c1}x}}&=\var{e1}\\ \var{c1}x &= \ln(\var{e1})\\ x &=\dfrac{\ln(\var{e1})}{\var{c1}}\\ \end{align}$

$x=$ {precround(ln(e1)/c1,2)}

Likningen $\simplify{{a2}e^(2*x)+{c2}e^(x)+{b2}=0}$ er en andregradslikning med $e^{x}$ som ukjent. Vi bruker da først abc-formelen for å finne $e^{x}$ :

$a=\var{a2},\;b=\var{c2},\;c=\var{b2}$

$e^{x}=\dfrac{\simplify{-{c2}}\pm\sqrt{\simplify[!collectNumbers,timesDot]{{c2}^{2}-4*{a2}*{b2}}}}{2\cdot\var{a2}}=\dfrac{\simplify{-{c2}}\pm\sqrt{\simplify{{c2}^2-4*{a2}*{b2}}}}{\simplify{2*{a2}}}$ .

Vi to løsninger for $e^{x}$ : $e^{x}= \var{x2[0]}\text{ og } e^{x}=\var{x2[1]}$ .

$e^{x}= \var{x2[0]}\Rightarrow x=\ln(\var{x2[0]})\approx\underline{\underline{\var{sol2[0]}}}$

$e^{x}= \var{x2[1]}\Rightarrow x=\ln(\var{x2[1]})\approx\underline{\underline{\var{sol2[1]}}}$

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This question is used in the following exam:

Logaritmer by Torris Bakke in UiT forkurs.

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Torris Bakke 1 month, 3 weeks ago

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Torris Bakke 2 months ago

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Logaritme- og eksponentiallikninger

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Torris Bakke 1 month, 3 weeks ago

Torris Bakke 1 month, 3 weeks ago

Torris Bakke 2 months ago

Error in variable testing condition

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Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`