I know the very basic principles of logs

[BruceW]

I guess everyone learns differently. I think that when I practice something like this, it gets easier the more times I have done it. But on the other had, I often do have a sort of 'click' moment. Like when I understand something so that I would be able to explain it in my own words, or with my own reasoning.

I think class notes are probably good enough. Maybe get a maths textbook if you wanted to read another perspective. Different people prefer different reasoning/explanations, so a textbook can be useful. I don't know any to recommend though. I did GCSE maths (being from the uk), so google them if you'd like.

I recommend practice, that's usually what works for me. Send me a message if you have a particular question. I think that if you try hard to understand these ones, then they might become one of your strengths in maths.

Good night, lloydowen. I'm guessing you're from the uk too? (since its pretty late for me too).

 

[lloydowen]

Hey, good afternoon! It's revision time again... :P I'm still stuck on that same question, and I refuse to move on unless I can do it, I need to be able to do these questions :(

 

[lloydowen]

What I've got is, from Marks little help, that if I do something to one side, its the opposite on the other? So, xlog(75.2)=3xlog(10.5)+5log(10.5) becomes -3xlog(10.5) + xlog(75.2)=+5log(10.5) Am I thinking the right way here? so then I would have 3xlog(10.5) as a common factor ?

EDIT I can't factorise x, it's confusing when its got all these logs inside grr.

 

[lloydowen]

For factorisation I've got this (which I'm sure is wrong, can someone confirm?)

x(log(75.2)-3log(10.5))=5log(10.5)

Doesn't seem right, on the last '3log(10.5)' part because wouldn't x end up being the variable after ? :S

 

[BruceW]

Your equation was xlog(75.2) = 3xlog(10.5) -5log(10.5). (I think you just typed it down wrong). But your method to get the next bit is correct. So, using your method, but the correct equation, becomes: -3xlog(10.5)+xlog(75.2)=-5log(10.5) From here you can factorise to get x. The logs are just numbers.

 

[BruceW]

For factorisation I've got this (which I'm sure is wrong, can someone confirm?)

x(log(75.2)-3log(10.5))=5log(10.5)

Doesn't seem right, on the last '3log(10.5)' part because wouldn't x end up being the variable after ? :S

This would be correct, but the right hand side of the equation should be negative of what you have written here.

Were you thinking it was wrong because the x was in-between the 3 and the log? Nope, that's fine to pull x out, because the order of multiplication doesn't matter.

 

[lloydowen]

This would be correct, but the right hand side of the equation should be negative of what you have written here.

How does it become negative? Sorry if that's a n00b question, trying to pick up on the principles of factorisation :P

 

[lloydowen]

Any feedback for that please?

x=\frac{log(75.2)-3log(10.5)}{-5log(10.5)}

Is that correct?

 

[lloydowen]

I tried it in a calculator and I didn't get teh correct answer :( where did I go wrong?

 

[BruceW]

The fraction is upside-down. You had xa=b, so you must divide both sides by a to get x on its own.

 

[BruceW]

How does it become negative? Sorry if that's a n00b question, trying to pick up on the principles of factorisation :P

It doesn't become negative, it always was negative. If you look back on the last page, the equation was xlog(75.2)=3xlog(10.5)-5log(10.5), but then on this page you copied it over as xlog(75.2)=3xlog(10.5)+5log(10.5) so you copied it over incorrectly (which happens a lot while doing maths, but less often if you check it).

 

[lloydowen]

It doesn't become negative, it always was negative. If you look back on the last page, the equation was xlog(75.2)=3xlog(10.5)-5log(10.5), but then on this page you copied it over as xlog(75.2)=3xlog(10.5)+5log(10.5) so you copied it over incorrectly (which happens a lot while doing maths, but less often if you check it).

Oh right, sorry It's hard to keep track of what I'm doing on forum. But what I had still worked out didn't work out to be the correct answer? What did I do wrong :(?

 

[BruceW]

You had x(log(75.2)-3log(10.5)) = -5log(10.5) (Which is correct). But from here, you wrote:
x= \frac{log(75.2)-3log(10.5)}{-5log(10.5)}
Which is not correct. Remember the rule of doing the same to each side.

So going from the correct equation, you divide on the left by (log(75.2)-3log(10.5)), so you must do the same on the right-hand side.

 

[lloydowen]

Not sure what you mean by that, I'm supposed to divide x(log(75.2)−3log(10.5))=−5log(10.5) to the right side too? (-5log(10.5)

 

[eumyang]

Not sure what you mean by that, I'm supposed to divide x(log(75.2)−3log(10.5))=−5log(10.5) to the right side too? (-5log(10.5)

But (log(75.2)-3log(10.5)) is what is being multiplied by x. So you do what BruceW said, and divide both sides by (log(75.2)-3log(10.5)):
x(log(75.2)-3log(10.5)) = -5log(10.5)
\frac{x(log(75.2)-3log(10.5))}{log(75.2)-3log(10.5)} = \frac{-5log(10.5)}{log(75.2)-3log(10.5)}
x= \frac{-5log(10.5)}{log(75.2)-3log(10.5)}

 

[BruceW]

Not sure what you mean by that, I'm supposed to divide x(log(75.2)−3log(10.5))=−5log(10.5) to the right side too? (-5log(10.5)

I'm not sure what you mean by this :) divide to the right side? I think you should practice the elementary algebra. Do you have class notes on that kind of stuff? For example, ax=b, then what is x?

P.S. eumyang has written out the answer to the question neatly.

 

[lloydowen]

I'm not sure what you mean by this :) divide to the right side? I think you should practice the elementary algebra. Do you have class notes on that kind of stuff? For example, ax=b, then what is x?

P.S. eumyang has written out the answer to the question neatly.

Yes but the answer that was written (Thank you so much btw) didn't get me the same answer in the answer book, and I'm unsure as to why :(

x=b/a... I can do things like that because they don't include numbers etc, but when numbers are involved I'm not sure why but I find it really difficult to re-arrange :( I'll look over some notes tomorrow in my lunch. Thanks for the help guys!

 

[eumyang]

Yes but the answer that was written (Thank you so much btw) didn't get me the same answer in the answer book, and I'm unsure as to why :(

Was the answer in the answer book in decimal form, or was the answer written as an expression involving logs? If the answer was in decimal form, and you are not getting the same answer, then possibly you're typing into the calculator wrong.

 

[lloydowen]

Sorry I've got the correct answer now, it's in decimal form and all I had to do was change the -5log to positive :):):) thanks

 

[BruceW]

x=b/a... I can do things like that because they don't include numbers etc, but when numbers are involved I'm not sure why but I find it really difficult to re-arrange :( I'll look over some notes tomorrow in my lunch. Thanks for the help guys!

I know exactly what you mean. I have the same trouble myself sometimes.

If there is a really big equation with numbers all over the place, I define algebraic constants, and write their values on one side of the page. Then I write the algebraic constants into my equation to make it look simpler, then I do the rearranging, then finally I put the actual numbers back in at the end.

For example, if I had the equation 54log(7)x+log(5)+ 87 \pi x = 0 Then I might choose to define a=54log(7), b=log(5), c= 87 \pi So then I would 'substitute in' my defined constants, so the equation now looks like ax+b +cx=0 So now it looks much simpler, and I would now rearrange the equation to x=\frac{-b}{a+c} And now that I am nearing the end of the question, I would write in the numbers for a,b,c to get x=\frac{-log(5)}{54log(7)+87}

So you see that I haven't really done anything special by defining these constants, but they just make it easier for me to do the rearranging.

 

[lloydowen]

Wow, that's actually a really good way to do it! I will try my best to remember all this for the test. I need to start practicing my re-arranging because my opinion is that if you remember the formula for the correct equation most of the rest will be re-arranging to make the subject of the formula and when you have a lot of numbers it does confuse me and I think it would confuse anyone, I will try and apply your method and see how it goes!

Practice makes perfect they say, Thanks for all your help everyone! Best forum commnuity I have probably ever come across! I will do my best to help others out too! :)

 

[BruceW]

good luck on your test. or 'ganbatte ne' as the japanese say. PF is the best forum community? that's pretty good. Yeah, this bunch are pretty tight. Probably because a large part of the forum is about helping to explain stuff.