Rearranging Logarithms involving Indices

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Elliott Fletcher 7 years, 9 months ago

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Created this as a copy of Combining Logarithm Rules to Solve Equations.

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

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Question statement

When logarithms involve indices we can rearrange them using the rule,

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

This can also be useful for removing integers from the front of logarithms.

Give any introductory information the student needs.

			Name	Type	Generated Value

p	integer	4
v	integer	7
q	number	2401
m	integer	20

			Name	Type	Generated Value

x1	list	[ 4, 4, 4, 5, 3, 4, 3, 2 ]
y1	integer	5
z1	list	List of 10 items
b1	integer	6
c	number	36
b4	integer	2
b	number	18
b2	number	16

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Value

xxxxxxxxxx
 
random(3..6)

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Description

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Generated value: `integer`

→ Used by:

Parts

Part a) Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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Part b) Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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Part c) Gap-fill

Gap-fill

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

Simplify the following expressions.

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

ii)

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

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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Elliott Fletcher 7 years, 9 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Hannah Aldous 7 years, 10 months ago

Hannah Aldous 7 years, 10 months ago

Error in variable testing condition

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Generated value: `integer`

Parts

Gap-fill

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c)

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Rearranging Logarithms involving Indices

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Taxonomy: mathcentre

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Elliott Fletcher 7 years, 9 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Hannah Aldous 7 years, 10 months ago

Hannah Aldous 7 years, 10 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

a)

b)

c)

Generated value: `integer`