Help with solving logarhythm problems

  • Thread starter Thread starter wasteofo2
  • Start date Start date
AI Thread Summary
To solve the logarithm problem log_c(√12), start by rewriting it as log_c(12)^(1/2) = (1/2)log_c(12). Using the properties of logarithms, log_c(12) can be expressed as log_c(2) + log_c(6), where log_c(6) can be further broken down into log_c(2) + log_c(3). Given log_c(2) = 1.214 and log_c(3) = 1.875, the values can be substituted to find log_c(12). The discussion emphasizes using logarithmic properties to simplify the problem effectively.
wasteofo2
Messages
477
Reaction score
2
How would I go about solving a problem like this?

Given log_c 3 = 1.875 and log_c 2 = 1.214 evaluate log_c {\sqrt{12}

What method would I have to use to solve a question like this?
 
Physics news on Phys.org
How about
\log_{c}\sqrt{12}=\log_{c}(12)^{\frac{1}{2}}=\frac{1}{2}\log_{c}12

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:
 
log_c 2 \sqrt 3 = log_c 2 + log_c \sqrt 3

you know what log_c 2 is

Thanks a lot for your kind words

NOTE: should be \sqrt 3 not \sqrt 2 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:
logarhythm = logarhythm

Take the antilog of both sides:
arhythm = arhythm

Replace hy = i[/tex]:<br /> arithm = arithm<br /> <br /> Take the log of both sides again and you&#039;re done: <br /> logarithm = logarithm <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:" />
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top