Help with solving logarhythm problems

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Homework Help Overview

The discussion revolves around solving logarithmic problems, specifically evaluating log_c of a square root expression using given logarithmic values.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss properties of logarithms, such as the logarithm of a square root and the addition of logarithms. There are attempts to manipulate the expression log_c(√12) using known logarithmic values.

Discussion Status

Some participants have provided guidance on how to approach the problem using logarithmic properties. There is acknowledgment of previous contributions, and a mix of interpretations is being explored without a clear consensus on the next steps.

Contextual Notes

There are mentions of specific logarithmic values provided in the problem, and some participants note corrections to earlier posts regarding the expressions used.

wasteofo2
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How would I go about solving a problem like this?

Given [tex]log_c 3 = 1.875[/tex] and [tex]log_c 2 = 1.214[/tex] evaluate [tex]log_c {\sqrt{12}[/tex]

What method would I have to use to solve a question like this?
 
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How about
[tex]\log_{c}\sqrt{12}=\log_{c}(12)^{\frac{1}{2}}=\frac{1}{2}\log_{c}12[/tex]

Take it from here.
Use the data & the properties of the "log" function.

Daniel.
 
courtigrad, following the previous advice dextericoby previously gave me... Your post was great, and your approach to the problem was fine. Don't delete your great posts! :smile:
 
[tex]log_c 2 \sqrt 3 = log_c 2 + log_c \sqrt 3[/tex]

you know what [tex]log_c 2[/tex] is

Thanks a lot for your kind words

NOTE: should be [tex]\sqrt 3[/tex] not [tex]\sqrt 2[/tex] for second part of addition. For some reason it will not let me change it.
 
Last edited:
wasteofo2 said:
Help with solving logarhythm problems
Start by setting it equal to itself:
[itex]logarhythm = logarhythm[/itex]

Take the antilog of both sides:
[itex]arhythm = arhythm[/itex]

Replace [itex]hy = i[/tex]:<br /> [itex]arithm = arithm[/itex]<br /> <br /> Take the log of both sides again and you're done: <br /> [itex]logarithm = logarithm[/itex] <br /> <br /> <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f61b.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":-p" title="Stick Out Tongue :-p" data-smilie="7"data-shortname=":-p" /> <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f631.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":eek:" title="Eek! :eek:" data-smilie="9"data-shortname=":eek:" /> <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f600.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":biggrin:" title="Big Grin :biggrin:" data-smilie="8"data-shortname=":biggrin:" /> <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f644.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":rolleyes:" title="Roll Eyes :rolleyes:" data-smilie="11"data-shortname=":rolleyes:" /> <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f644.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":rolleyes:" title="Roll Eyes :rolleyes:" data-smilie="11"data-shortname=":rolleyes:" />[/itex]
 

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