Numerical skills - Solving Logarithms - Numbas at mathcentre.ac.uk

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joshua boddy

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8 years, 9 months ago

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joshua boddy 8 years, 9 months ago

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Created this as a copy of Solving Logarithms .

Name	Status	Author	Last Modified
Solving Logarithms	Ready to use	joshua boddy	01/06/2016 09:47
Numerical skills - Solving Logarithms	Ready to use	Jinhua Mathias	28/08/2024 12:15
Numerical skills - Solving Logarithms summative	Ready to use	Jinhua Mathias	28/08/2024 12:15
Numerical skills Solving Logarithms	draft	Jinhua Mathias	27/08/2024 16:37
Solving Logarithms for Geosciences	Ready to use	Jinhua Mathias	27/08/2024 16:37
Andrew's copy of Numerical skills Solving Logarithms	draft	Andrew Dunbar	30/06/2017 22:04
Solving Logarithms for Geosciences	draft	Xiaodan Leng	11/07/2019 05:11
Solving Logarithms	draft	Xiaodan Leng	11/07/2019 06:36

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Question statement

Solve these equations for x

Give any introductory information the student needs.

			Name	Type	Generated Value

a1	number	200
a2	integer	10
a3	rational	5
a4	integer	3
b1	integer	4
b2	integer	2
b3	integer	3
b4	integer	9
c1	integer	2
c2	integer	6
c3	integer	9
c4	integer	4
c5	integer	2
a	integer	6
b	integer	9
d1	integer	3
d2	integer	3
d3	integer	54
d4	integer	15

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
2a2^((a4)-1)

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `number`

200

← Depends on:

→ Used by:

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

Parts

Part a) Gap-fill
1. Gap Number entry
  Add
Part b) Gap-fill
1. Gap Number entry
  Add

Gap-fill

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

$\log_\var{a2}\var{a3}+\log_\var{a2}x = \var{a4}$

$x=$ Unnamed gap

Advice

We want to solve for x. After simplifying, in each case we end up with $\log_a{f(x)} = b$ , so we raise both sides as a power of $a$ to get $a^{\log_a{f(x)}} = a^b$ which simplifies (by laws of logarithms) to $f(x)=a^b$ . We then solve for x accordingly.

Use of the laws of logarithms is crucial here:

$\log{a} + \log{b} = \log{ab}$

$\log{a} - \log{b} = \log{\frac{a}{b}}$

$\log{a^n} = n\log{a}$

Give a worked solution to the whole question.

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There was an error which means the tests can't run:

Part	Test	Passed?
Gap-fill
Gap-fill		Hasn't run yet
Number entry
Number entry		Hasn't run yet
Gap-fill
Gap-fill		Hasn't run yet
Number entry
Number entry		Hasn't run yet

This question is used in the following exam:

Foundation Maths end of module test practice paper by Jinhua Mathias in Core Foundation Maths.

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Numerical skills - Solving Logarithms

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Taxonomy: mathcentre

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joshua boddy 8 years, 9 months ago

Deactivated user 9 years, 4 months ago

Error in variable testing condition

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Generated value: `number`

Parts

Gap-fill

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Numerical skills - Solving Logarithms

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

joshua boddy 8 years, 9 months ago

Deactivated user 9 years, 4 months ago

Error in variable testing condition

Error

Warning

Generated value: number

Parts

Gap-fill

Generated value: `number`