Ugur's copy of Logarithms1

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Ugur Efem

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3 years, 3 months ago

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Ugur Efem 3 years, 3 months ago

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Created this as a copy of Logarithms1.

Name	Status	Author	Last Modified
SFY0004 Logarithms1	draft	Bill Foster	06/08/2020 08:42
Logarithms1	Ready to use	Wan Mekwi	13/10/2020 13:20
Ugur's copy of Logarithms1	Ready to use	Ugur Efem	14/01/2022 15:08

There are 14 other versions that do you not have access to.

Question statement

Answer the following question on logarithms.

Give any introductory information the student needs.

			Name	Type	Generated Value

c	number	102
b	integer	7
a_3	integer	6
a_2	integer	6
a_1	integer	17
r_1	integer	2
r_2	integer	3
tol	number	0.0001
ans	number	2.3768

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
a_1*a_3^r_2/a_2^r_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`

102

← Depends on:

→ Used by:

This variable doesn't seem to be used anywhere.

Parts

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

Gap-fill

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

Find a fraction or integer $c$ such that:

$\log_{\var{b}}(\var{a_1})-\var{r_1}\log_{\var{b}}(\var{a_2})+\var{r_2}\log_{\var{b}}(\var{a_3})=\log_{\var{b}}(c)$

$c=\;$ Gap 0..

Input $c$ as an integer or as a fraction and not as a decimal.

Now calculate $\log_{\var{b}}(c)$ to 4 decimal places:

$\log_{\var{b}}(c)=\;$ Gap 1..

Advice

The rules for combining logs are

$\begin{eqnarray*}&1.& \log_b(ac)&=&\log_b(a)+\log_b(c)\\ \\ &2.& \log_b\left(\frac{a}{c}\right)&=&\log_b(a)-\log_b(c)\\ \\ &3.& \log_b(a^r)&=&r\log_b(a) \end{eqnarray*}$

We see that:

$\begin{eqnarray*}\log_{\var{b}}(\var{a_1})-\var{r_1}\log_{\var{b}}(\var{a_2})+\var{r_2}\log_{\var{b}}(\var{a_3})&=&\log_{\var{b}}(\var{a_1})-\log_{\var{b}}(\var{a_2}^{\var{r_1}})+\log_{\var{b}}(\var{a_3}^{\var{r_2}})\mbox{ using 3.}\\&=&\log_{\var{b}}(\var{a_1})-\log_{\var{b}}(\var{a_2^r_1})+\log_{\var{b}}(\var{a_3^r_2})\\&=&\log_{\var{b}}(\var{a_1}\times \var{a_3^r_2})-\log_{\var{b}}(\var{a_2^r_1}) \mbox{ using 1.}\\&=&\log_{\var{b}}\left(\frac{\var{a_1}\times \var{a_3^r_2}}{\var{a_2^r_1}}\right) \mbox{ using 2.}\\&=&\log_{\var{b}}\left(\frac{\var{a_1*a_3^r_2}}{\var{a_2^r_1}}\right)\\&=&\log_{\var{b}}\left(\simplify[all,fractionnumbers]{{a_1*a_3^r_2}/{a_2^r_1}}\right)\mbox{ on cancelling common factors}.\end{eqnarray*}$

Hence $\displaystyle c=\simplify[all,fractionnumbers]{{a_1*a_3^r_2}/{a_2^r_1}}$ .

To calculate $\displaystyle \log_{\var{b}}\left(\simplify[all,fractionnumbers]{{a_1*a_3^r_2}/{a_2^r_1}}\right)$ to 4 decimal places we use the fact that for any positive base $b$ :

$\log_b(c)=\frac{\ln(c)}{\ln(b)}=\frac{\log_{10}(c)}{\log_{10}(b)}$

and we can use either of the log functions, $\ln$ or $\log_{10}$ on our calculators to find the value.

Using $\ln$ we find:

$\log_{\var{b}}\left(\simplify[all,fractionnumbers]{{a_1*a_3^r_2}/{a_2^r_1}}\right)=\frac{\ln\left(\simplify[all,fractionnumbers]{{a_1*a_3^r_2}/{a_2^r_1}}\right)}{\ln(\var{b})}=\var{ans}$ to 4 decimal places.

Give a worked solution to the whole question.

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

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Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

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Ugur Efem 3 years, 3 months ago

Ugur Efem 3 years, 3 months ago

Ugur Efem 3 years, 4 months ago

Error in variable testing condition

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Generated value: `number`

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Ugur's copy of Logarithms1

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Ugur Efem 3 years, 3 months ago

Ugur Efem 3 years, 3 months ago

Ugur Efem 3 years, 4 months ago

Error in variable testing condition

Error

Warning

Generated value: number

Parts

Gap-fill

Generated value: `number`