Logarithms: Solving equations 2

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Bill Foster

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Bill Foster 55 years, 2 months ago

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Name	Status	Author	Last Modified
Logarithms: Solving equations 2	draft	Bill Foster	06/08/2020 08:42
Logarithms: Solving equations 2	Ready to use	Julie Crowley	26/07/2016 12:53
Regina's copy of Logarithms: Solving Equations 2	draft	Regina Kobus	07/11/2017 01:22
Regina's copy of Logarithms: Solving Equations 2	draft	Regina Kobus	07/11/2017 01:43
David's copy of Logarithms: Solving equations 2	draft	David Martin	30/07/2018 14:24
Simon's copy of Logarithms: Solving equations 2	draft	Simon Thomas	04/03/2019 11:28
Logarithms: Solving equations 2	draft	Xiaodan Leng	11/07/2019 05:47
Logarithms: Solving equations 2	Ready to use	Wan Mekwi	13/10/2020 13:20

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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	3
c	number	5
b	integer	-1
d	integer	1
s	integer	-1
sol2	number	-2
sol1	number	7

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Generated value: `integer`

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c
sol2

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"Unnamed part" - prompt

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$2\log_{\var{a}}(\simplify{x+{b}})- \log_{\var{a}}(\simplify{(x+{c})})=\var{d}$

$x=\;$ 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 three rules for logs :

1. $n\log_a(m)=\log_a(m^n)$

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

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

Using rule 1 we get

$2\log_{\var{a}}(\simplify{x+{b}})- \log_{\var{a}}(\simplify{(x+{c})})=\log_{\var{a}}((\simplify{x+{b}})^2)- \log_{\var{a}}(\simplify{(x+{c})})$
Using rule 2 gives

$\log_{\var{a}}(\simplify{(x+{b})^2})- \log_{\var{a}}(\simplify{(x+{c})})=\log_{\var{a}}\left(\simplify{(x+{b})^2/(x+{c})}\right)$
So the equation to solve becomes:

$\log_{\var{a}}\left(\simplify{(x+{b})^2/(x+{c})}\right)=\var{d}$
and using rule 3 this gives:

$\begin{eqnarray} \simplify{(x+{b})^2/(x+{c})}&=&{\var{a}}^{\var{d}}\Rightarrow\\ \simplify{(x+{b})^2}&=&{\var{a}}^{\var{d}}(\simplify{x+{c}})=\simplify{{a^d}(x+{c})}\Rightarrow\\ \simplify{x^2+{2*b-a^(d)}x+{b^2-a^(d)*c}}&=&0 \end{eqnarray}$
Solving this quadratic we get two solutions:

$x=\var{sol1}$ and $x=\var{sol2}$

We should check that these solutions gives positive values for $\simplify{x+{b}}$ and $\simplify{x+{c}}$ as otherwise the logs are not defined.

The value $x=\var{sol1}$ gives:

Substituting this value for $x$ into $\log_{\var{a}}(\simplify{x+{b}})$ we get $\log_{\var{a}}(\simplify{{2*a^d}})$ so OK.

Substituting this value for $x$ into $\log_{\var{a}}(\simplify{x+{c}})$ we get $\log_{\var{a}}(\simplify{{4*a^d}})$ so OK.

Hence $x=\var{sol1}$ is a solution to our original equation.

The value $x=\var{sol2}$ gives:

Substituting this value for $x$ into $\log_{\var{a}}(\simplify{x+{b}})$ we get $\log_{\var{a}}(\simplify{{-a^d}})$ so NOT OK.

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

Hence $x=\var{sol2}$ is a solution to our original equation as $\log_{\var{a}}(\simplify{x+{b}})$ is not defined for this value of $x$ .

So there is only one solution $x=\var{sol1}$ .

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Bill Foster 55 years, 2 months ago

Error in variable testing condition

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Generated value: `integer`

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Logarithms: Solving equations 2

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Bill Foster 55 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`