View Full Version : help with logarithmic equations
chirumu
Oct16-11, 12:37 AM
1. The problem statement, all variables and given/known data
http://img225.imageshack.us/img225/6162/unledsxy.png
2. Relevant equations
logb(x)-logb(y) = logb(x/y)
logb(x^n) = nlogb(x)
3. The attempt at a solution
i am stuck on what to do because of the 4-2-2 portion. I've attempted it multiple times to get the solution (-4) but i just dont get how to do it because i keep getting the answer wrong.
I guess i dont understand how to combine it into one log this is my attempt nontheless:
log3(4-2)-2log3(6)
log3(2)-log3(12)
log3(2/12)
log3(x) = (2/12)
:/ i'm not sure what to do because i dont know how to treat the 4-2-2 :/ i know i cant combine the -2 where i want..
actually no wait, i've just had a mindwave. Would i convert 2 into a log as well.
I will try that now.
What's the question you're working on?
There is no question given in your post.
1. The problem statement, all variables and given/known data
http://img225.imageshack.us/img225/6162/unledsxy.png
I am not certain what your question is, like Sammy, here. However, I tried to simplify the equation (quoted above) and it brought me to -4. I understand this is the answer you were looking for? If yes, read down:
Step 1: Figure out how to change "2" to a log of base 3.
Step 2: If you've managed to do step 1, then think of your laws of logarithm and then, indices.
Now that I've taken a closer look at your attempt to this problem, I see that you don't seem to know your laws of logarithm.
I'll try to do a breakdown of a few of the operations and hopefully, things will make more sense.
What you did was:
2\log_{3}6 = log_{3}(6\times2)
Here's how it should have been manipulated:
2\log_{3}6 = log_{3}(6^2)= log_{3}36
Also, in this situation:
2 = log_{3}(3^2)
I would urge you to review your laws of logarithm because it seems to me that this is not very clear to you. I think the website math2.org has some good explanations.
chirumu
Oct17-11, 08:33 AM
thankyou for your help :)
i initially skipped this one and moved onto the others and after having done a few more its become more clear to me.
in the end:
log3(4)-2-2log3(6)
given a^c=b and loga(b)=c
to change 2 to base 3: log3(9) =2
so
log3(4)-log3(9)-2log3(6)
log3(4)-log3(9)-log3(6^2)
log3(4)-log3(9)-log3(36)
log3(4/9/36) = x
3(x) = (4/9/36)
x = -4
however i couldnt figure out -4 without a calculator :I i'm trying to stop using my calculator as much because its the devil. I've normally been able to solve the power, but in this instance the power is negative and the answer (4/9/36) is 0.01234... and i wasnt sure. Upon inspection given the fact that (4/9/36) IS 0.01234... i could kind of guess the power would be negative, but i would not of been able to figure it was -4. Could you offer any advice on this.
I am not certain what your question is, like Sammy, here. However, I tried to simplify the equation (quoted above) and it brought me to -4. I understand this is the answer you were looking for? If yes, read down:
The image in the OP is NOT an equation. An equation ALWAYS has "=" in it somewhere.
thankyou for your help :)
i initially skipped this one and moved onto the others and after having done a few more its become more clear to me.
in the end:
log3(4)-2-2log3(6)
given a^c=b and loga(b)=c
to change 2 to base 3: log3(9) =2
so
log3(4)-log3(9)-2log3(6)
log3(4)-log3(9)-log3(6^2)
log3(4)-log3(9)-log3(36)
log3(4/9/36) = x
3(x) = (4/9/36)
x = -4
Your first post did not make this clear, but apparently you are supposed to simplify the given expression. When you rewrite an expression in a different form, use = to indicate that the different forms are equal.
You do not need to introduce a variable. Your work should look something like this:
log3(4) - 2 - 2log3(6)
= log3(4) - log3(9) - 2log3(6)
= log3(4/9) - 2log3(6)
= log3(4/9) - log3(36)
= log3(4/((9*36)
If you continue simplifying, you can get this down to a final result of -4.
however i couldnt figure out -4 without a calculator :I i'm trying to stop using my calculator as much because its the devil. I've normally been able to solve the power, but in this instance the power is negative and the answer (4/9/36) is 0.01234... and i wasnt sure. Upon inspection given the fact that (4/9/36) IS 0.01234... i could kind of guess the power would be negative, but i would not of been able to figure it was -4. Could you offer any advice on this.
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.