Logarithms: Solving equations 1

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Wan Mekwi

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4 years, 5 months ago

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Wan Mekwi 4 years, 5 months ago

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Wan Mekwi 4 years, 5 months ago

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Created this as a copy of Logarithms: Solving equations 1.

Name	Status	Author	Last Modified
Logarithms: Solving equations 1	draft	Bill Foster	06/08/2020 08:42
Logarithms: Solving equations 1	Ready to use	Julie Crowley	26/07/2016 12:53
David's copy of Logarithms: Solving equations 1	draft	David Martin	30/07/2018 14:24
Simon's copy of Logarithms: Solving equations 1	draft	Simon Thomas	04/03/2019 11:22
Logarithms: Solving equations 1	draft	Xiaodan Leng	11/07/2019 05:47
Logarithms: Solving equations 1	Ready to use	Wan Mekwi	13/10/2020 13:27

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Question statement

Solve the following equation for $x$ .

Input your answer as a fraction or an integer as appropriate and not as a decimal.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	2
c	integer	14
b	integer	16
d	integer	1

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Generated value: `integer`

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"Unnamed part" - prompt
"Unnamed part" → "Unnamed gap" - Correct answer

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1. Gap Mathematical expression
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$\log_{\var{a}}(x+\var{b})- \log_{\var{a}}(\simplify{(x+{c})})=\var{d}$

$x=\;$ Unnamed gap

If you want help in solving the equation, click on Show steps. If you do so then you will lose 1 mark.

Input all numbers as fractions or integers and not as decimals.

Advice

We use the following two rules for logs :

1. $\log_a(b)-\log_a(c)=\log_a(b/c)$

2. $\log_a(p)=r \Rightarrow p=a^r$

Using rule 1 we get

$\log_{\var{a}}(x+\var{b})- \log_{\var{a}}(\simplify{(x+{c})})=\log_{\var{a}}\left(\simplify{(x+{b})/(x+{c})}\right)$
So the equation to solve becomes:

$\log_{\var{a}}\left(\simplify{(x+{b})/(x+{c})}\right)=\var{d}$
and using rule 2 this gives:

$\begin{eqnarray} \simplify{(x+{b})/(x+{c})}&=&{\var{a}}^{\var{d}}\Rightarrow\\ x+\var{b}&=&{\var{a}}^{\var{d}}(x+\var{c})=\simplify{{a^d}}(x+\var{c})\Rightarrow\\ \simplify{{a^d-1}x}&=&\simplify[std]{{b}-{c}*{a^d}={b-c*a^d}}\Rightarrow\\ x&=&\simplify{{b-c*a^d}/{a^d-1}} \end{eqnarray}$
We should check that this solution gives positive values for

$x+\var{b}$ and

$\simplify{x+{c}}$ as otherwise the logs are not defined.

Substituting this value for $x$ into $\log_{\var{a}}(x+\var{b})$ we get $\log_{\var{a}}(\simplify{({b-c }{a^d})/{a^d-1}})$ so OK.

For $\log_{\var{a}}(\simplify{x+{c}})$ we get on substituting for $x$ , $\log_{\var{a}}(\simplify{({b-c })/{a^d-1}})$ so OK.

Hence the value we found for $x$ is a solution to the original equation.

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Taxonomy: mathcentre

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Wan Mekwi 4 years, 5 months ago

Wan Mekwi 4 years, 5 months ago

Wan Mekwi 4 years, 5 months ago

Error in variable testing condition

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Generated value: `integer`

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Logarithms: Solving equations 1

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Wan Mekwi 4 years, 5 months ago

Wan Mekwi 4 years, 5 months ago

Wan Mekwi 4 years, 5 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`