Logarithms1

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From users who are members of Introduction to Calculus :

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

Gave some feedback: Ready to use

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Wan Mekwi 4 years, 5 months ago

Created this as a copy of SFY0004 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	0.0031828704
b	integer	6
a_3	integer	2
a_2	integer	24
a_1	integer	11
r_1	integer	3
r_2	integer	2
tol	number	0.0001
ans	number	-3.2091

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Name

Data type

Value

xxxxxxxxxx
 
a_1*a_3^r_2/a_2^r_1

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Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `number`

0.0031828704

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Parts

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

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This question is used in the following exams:

Exponentials and Logarithms by Wan Mekwi in Introduction to Calculus.
erste sammlung by Karin Buhler in Karin's workspace.
Week 17 logarithm by Izabela Walczak in Izabela's workspace.

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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

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Logarithms1

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

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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: number

Parts

Gap-fill

Generated value: `number`