cormac's copy of Ida's copy of Rearranging Logarithms involving Indices

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cormac breen 6 years, 5 months ago

Created this as a copy of Ida's copy of Rearranging Logarithms involving Indices.

Name	Status	Author	Last Modified
Combining Logarithm Rules to Solve Equations	Ready to use	Hannah Aldous	20/11/2019 14:34
Using Laws for Addition and Subtraction of Logarithms	Ready to use	Hannah Aldous	20/11/2019 14:44
Rearranging Logarithms involving Indices	Ready to use	Hannah Aldous	20/11/2019 14:44
Addition and Subtraction of Logarithms	Ready to use	Adrian Jannetta	16/08/2023 14:56
MATH6058 Addition and Subtraction of Logarithms	draft	Catherine Palmer	09/10/2017 11:30
Raphael's copy of Rearranging Logarithms involving Indices	draft	Raphael Stuhlmeier	27/11/2017 16:06
Addition and Subtraction of Logarithms	draft	Ida Landgärds	16/04/2018 10:43
Ida's copy of Rearranging Logarithms involving Indices	Ready to use	Ida Landgärds	09/05/2018 06:55
spørsmål 3	draft	Ida Landgärds	02/12/2020 15:42
Jonathan's copy of Using Laws for Addition and Subtraction of Logarithms	draft	Jonathan Kress	07/06/2018 01:10
Jeanne's copy of Using Laws for Addition and Subtraction of Logarithms	draft	Jeanne Parmentier	20/07/2018 09:31
Using Laws for Addition and Subtraction of Logarithms [L6 Randomised]	Needs to be tested	Matthew James Sykes	25/07/2018 11:08
Rearranging Logarithms involving Indices [L6]	draft	Abbi Mullins	24/07/2018 14:39
logarithms addition/subtraction [L6]	draft	Abbi Mullins	24/07/2018 14:49
NUMBAS-Logarithms (laws)	Ready to use	Katy Dobson	30/07/2018 10:39
David's copy of Using Laws for Addition and Subtraction of Logarithms	Ready to use	David Martin	21/09/2018 11:50
David's copy of Combining Logarithm Rules to Solve Equations	draft	David Martin	30/07/2018 14:34
David's copy of Rearranging Logarithms involving Indices	Ready to use	David Martin	17/10/2019 07:25
Joël's copy of Using Laws for Addition and Subtraction of Logarithms	draft	Jo Cohen	02/12/2020 15:42
Blathnaid's copy of spørsmål 3	draft	Blathnaid Sheridan	02/12/2020 15:42
cormac's copy of Ida's copy of Rearranging Logarithms involving Indices	draft	cormac breen	05/11/2018 13:37
cormac's copy of Ida's copy of Rearranging Logarithms involving Indices	draft	cormac breen	05/11/2018 13:38
cormac's copy of Ida's copy of Rearranging Logarithms involving Indices	draft	cormac breen	05/11/2018 13:54
Justin's copy of Using Laws for Addition and Subtraction of Logarithms	draft	Justin McGuinness	14/05/2019 14:55
Maria's copy of Rearranging Logarithms involving Indices	draft	Maria Aneiros	11/06/2019 05:32
Maria's copy of Using Laws for Addition and Subtraction of Logarithms	draft	Maria Aneiros	11/06/2019 05:30
Maria's copy of Combining Logarithm Rules to Solve Equations	draft	Maria Aneiros	25/05/2019 08:02
Using Laws for Addition and Subtraction of Logarithms	draft	Xiaodan Leng	10/07/2019 23:28
Logarithms (laws)	draft	Xiaodan Leng	10/07/2019 23:57
Using Laws for Addition and Subtraction of Logarithms	draft	Xiaodan Leng	11/07/2019 00:00
Addition and Subtraction of Logarithms	draft	Xiaodan Leng	11/07/2019 00:15
Rearranging Logarithms involving Indices	draft	Xiaodan Leng	11/07/2019 00:16
Combining Logarithm Rules to Solve Equations	draft	Xiaodan Leng	11/07/2019 01:39
Katy's copy of Combining Logarithm Rules to Solve Equations	draft	Katy Dobson	03/09/2019 08:29
Katy's copy of Using Laws for Addition and Subtraction of Logarithms	draft	Katy Dobson	03/09/2019 08:29
Musa's copy of 3 Addition and Subtraction of Logarithms	draft	Musa Mammadov	27/06/2023 01:29
Rearranging Logarithms involving Indices	Ready to use	Wan Mekwi	13/10/2020 13:19
Combining Logarithm Rules to Solve Equations	Ready to use	Wan Mekwi	13/10/2020 13:22

There are 84 other versions that do you not have access to.

Question statement

Using the rules of logarithims, simplfy the following expressions

Give any introductory information the student needs.

			Name	Type	Generated Value

p	integer	5
v	integer	6
q	number	7776
m	integer	12

			Name	Type	Generated Value

x1	list	[ 3, 5, 4, 3, 3, 4, 4, 4 ]
y1	integer	2
z1	list	List of 10 items
b1	integer	8
c	number	512
b4	integer	3
b	number	256
b2	number	254

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(3..6)

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: `integer`

→ Used by:

Parts

Gap-fill
1. Gap Gap 0. Mathematical expression
  Add
2. Gap Gap 1. Mathematical expression
  Add

Gap-fill

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

$\log_a(\var{x1[4]^z1[1]})+\log_a(\var{x1[4]^z1[0]})=$ Gap 0. $\log_a(\var{x1[4]})$

ii)

$\log_a(\var{x1[5]^z1[1]})+\log_a(\var{x1[5]^z1[2]})-\log_a(\var{x1[5]^z1[4]})=$ Gap 1. $\log_a(\var{x1[5]})$

Advice

a)

We need to use the rule

$k\log_a(x)=\log_a(x^k)\text{.}$

Subsituting in our values for $x$ and $k$ gives

$\var{x1[3]}\log_a(\var{z1[0]})=\log_a(\var{z1[0]^x1[3]})\text{.}$

ii)

We need to use the rule

$k\log_a(x)=\log_a(x^k)\text{.}$

Subsituting in our values for $x$ and $k$ gives

$\var{x1[1]}\log_a(\var{z1[1]})=\log_a(\var{z1[1]^x1[1]})\text{.}$

b)

The rule for indices in logarithms also works the other way around,

$\log_a(x^k)=k\log_a(x)\text{.}$

We can use this to rearrange our expression by substituting in values for $x$ and $k$ .

$\begin{align} \log_a(\var{x1[3]^z1[5]})&=k\log_a(\var{x1[3]})\\ \var{x1[3]^z1[5]}&=\var{x1[3]}^k\\ \var{x1[3]^z1[5]}&=\var{x1[3]}^\var{z1[5]}\\ k&=\var{z1[5]}\\ \log_a(\var{x1[3]^z1[5]})&=\var{z1[5]}\log_a(\var{x1[3]}) \end{align}$

ii)

As with i) we can use the rule

$\log_a(x^k)=k\log_a(x)\text{.}$

We can use this to rearrange our expression by substituting in values for $x$ and $k$ .

$\begin{align} \log_a(\var{x1[5]^z1[6]})&=k\log_a(\var{x1[5]})\\ \var{x1[5]^z1[6]}&=\var{x1[5]}^k\\ \var{x1[5]^z1[6]}&=\var{x1[5]}^\var{z1[6]}\\ k&=\var{z1[6]}\\ \log_a(\var{x1[5]^z1[6]})&=\var{z1[6]}\log_a(\var{x1[5]}) \end{align}$

c)

From the structure of this question we can tell that the answer can be written in the form $k\log_a(\var{x1[3]})$ , meaning all of the values in the expression

$\log_a(\var{x1[3]^z1[2]})+\log_a(\var{x1[3]})$

can be written in the form $k\log_a(\var{x1[3]})$ .

If we look at each log individually we can make sure they all take this form.

$\begin{align} \log_a(\var{x1[3]^z1[2]})&=k\log_a(\var{x1[3]})\\ \var{x1[3]^z1[2]}&=\var{x1[3]}^k\\ \var{x1[3]^z1[2]}&=\var{x1[3]}^\var{z1[2]}\\ k&=\var{z1[2]}\\ \log_a(\var{x1[3]^z1[2]})&=\var{z1[2]}\log_a(\var{x1[3]}) \end{align}$

We can now write our expression as

$\begin{align} \log_a(\var{x1[3]^z1[2]})+\log_a(\var{x1[3]})&=\var{z1[2]}\log_a(\var{x1[3]})+\log_a(\var{x1[3]})\\ &=\var{z1[2]+1}\log_a(\var{x1[3]})\text{.} \end{align}$

ii)

From this question we know our answer is written in the form $k\log_a(\var{x1[4]})$ , meaning all of the values in the expression

$\log_a(\var{x1[4]^z1[1]})+\log_a(\var{x1[4]^z1[0]})$

can be written in the form $k\log_a(\var{x1[4]})$ .

If we look at each log individually we can make sure they all take this form.

$\begin{align} \log_a(\var{x1[4]^z1[1]})&=k\log_a(\var{x1[4]})\\ \var{x1[4]^z1[1]}&=\var{x1[4]}^k\\ \var{x1[4]^z1[1]}&=\var{x1[4]}^\var{z1[1]}\\ k&=\var{z1[1]}\\ \log_a(\var{x1[4]^z1[1]})&=\var{z1[1]}\log_a(\var{x1[4]}) \end{align}$

$\begin{align} \log_a(\var{x1[4]^z1[0]})&=k\log_a(\var{x1[4]})\\ \var{x1[4]^z1[0]}&=\var{x1[4]}^k\\ \var{x1[4]^z1[0]}&=\var{x1[4]}^\var{z1[0]}\\ k&=\var{z1[0]}\\ \log_a(\var{x1[4]^z1[0]})&=\var{z1[0]}\log_a(\var{x1[4]}) \end{align}$

We can now write our expression as

$\begin{align} \log_a(\var{x1[4]^z1[1]})+\log_a(\var{x1[4]^z1[0]})&=\var{z1[1]}\log_a(\var{x1[4]})+\var{z1[0]}\log_a(\var{x1[4]})\\ &=\var{z1[1]+z1[0]}\log_a(\var{x1[4]})\text{.} \end{align}$

iii)

From this question we know our answer is written in the form $k\log_a(\var{x1[5]})$ , meaning all of the values in the expression

$\log_a(\var{x1[5]^z1[1]})+\log_a(\var{x1[5]^z1[2]})-\log_a(\var{x1[5]^z1[4]})$

can be written in the form $k\log_a(\var{x1[5]})$ .

If we look at each log individually we can make sure they all take this form.

$\begin{align} \log_a(\var{x1[5]^z1[1]})&=k\log_a(\var{x1[5]})\\ \var{x1[5]^z1[1]}&=\var{x1[5]}^k\\ \var{x1[5]^z1[1]}&=\var{x1[5]}^\var{z1[1]}\\ k&=\var{z1[1]}\\ \log_a(\var{x1[5]^z1[1]})&=\var{z1[1]}\log_a(\var{x1[5]}) \end{align}$

$\begin{align} \log_a(\var{x1[5]^z1[2]})&=k\log_a(\var{x1[5]})\\ \var{x1[5]^z1[2]}&=\var{x1[5]}^k\\ \var{x1[5]^z1[2]}&=\var{x1[5]}^\var{z1[2]}\\ k&=\var{z1[2]}\\ \log_a(\var{x1[5]^z1[2]})&=\var{z1[2]}\log_a(\var{x1[5]}) \end{align}$

$\begin{align} \log_a(\var{x1[5]^z1[4]})&=k\log_a(\var{x1[5]})\\ \var{x1[5]^z1[4]}&=\var{x1[5]}^k\\ \var{x1[5]^z1[4]}&=\var{x1[5]}^\var{z1[4]}\\ k&=\var{z1[4]}\\ \log_a(\var{x1[5]^z1[4]})&=\var{z1[4]}\log_a(\var{x1[5]}) \end{align}$

We can now write our expression as

$\begin{align} \log_a(\var{x1[5]^z1[1]})+\log_a(\var{x1[5]^z1[2]})-\log_a(\var{x1[5]^z1[4]})&=\var{z1[1]}\log_a(\var{x1[5]})+\var{z1[0]}\log_a(\var{x1[5]})-\var{z1[4]}\log_a(\var{x1[5]})\\ &=\var{z1[1]+z1[2]-z1[4]}\log_a(\var{x1[5]})\text{.} \end{align}$

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cormac's copy of Ida's copy of Rearranging Logarithms involving Indices

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cormac breen 6 years, 5 months ago

Error in variable testing condition

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Generated value: `integer`

Parts

Gap-fill

a)

b)

c)

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cormac's copy of Ida's copy of Rearranging Logarithms involving Indices

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

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cormac breen 6 years, 5 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

a)

b)

c)

Generated value: `integer`