I know the very basic principles of logs

[lloydowen]

Homework Statement

I'm new to logs. I know the very basic principles of logs, unfortunately my notes are not helping my on this occasion. What I have so far is Solve the value 'x'

Homework Equations

6^(x+3)=56.34

The Attempt at a Solution

Here's what I have done, and I'm not sure where to go from here so could someone explain?

log6^(x+3)=log5.34
(x+3)log6=log56.34
(x+3)log6=1.751
------

Not sure if I've done it correctly, and I'm not sure where to go from next :/

If you have seen me asking a lot of questions tonight, I have an exam next week and I'm just getting ready for it :)

 

[micromass]

Good. Now you have an equation of the form

(x+3)b=c

Solve for x.

 

[lloydowen]

(x+3)b=c

Solve for x.

So, the next line would be xlog6 +3log6 = 1.751

Yes or no?

 

[BruceW]

I'm guessing on your first line, you meant log56.34 (since that's what you put on the next line, which is correct).

You have got it right so far. From here, its simple rearrangement to get x.

Edit: Sorry, I started writing this before micromass posted, so I am a little behind here.

yes, that would be the next line (although you could instead have divided both sides by b), but you get the right answer either way.

 

[micromass]

So, the next line would be xlog6 +3log6 = 1.751

Yes or no?

Yes.

 

[Mark44]

Here's what I have done, and I'm not sure where to go from here so could someone explain?

log6^(x+3)=log5.34
(x+3)log6=log56.34
(x+3)log6=1.751
------

As already noted, the first line above has a typo.

So, the next line would be xlog6 +3log6 = 1.751

Yes or no?

Yes.

As an alternative, starting from (x + 3)log(6) = log(56.34),
you could divide both sides by log(6) to get x + 3 = log(56.34)/log(6), and you can solve for x from there.

Note that put back in log(56.34) instead of using your value of 1.751. You'll get a more precise result if you DON'T put in rounded intermediate values. If you use the full values that your calculator gives you, you'll get answers that don't have built-in inaccuracies.

 

[lloydowen]

So, on that last line I done... xlog6 + 3log6

Do I (on a calculator) type log6= then Ans*3... and then add it to log6 ? I'm confused now :(

 

[lloydowen]

Any help guys? I'm real stuck ! Can't move on from here...

 

[lloydowen]

So I couldn't do multiplication method so I went for division.. Here's what I got.

(x+3=1.751)/log6

Uhh doesn't look right but is it ?

 

[micromass]

I really don't know what is bothering you so much. Can you solve

2(x+3)=5

??

The equation you list is the same thing but with different numbers.

 

[lloydowen]

So I think I have a final answer, I kept simplifying down. after that previous line i got (x=1.751)/log6 (-3)

Which then gave me a final answer of -0.75 to 2dp

 

[lloydowen]

New Question!

So I got another Question because I didn't full click with that last one..

75.2^x = 10.5 ^(3x-5)What I got so far...

log75.2^x=log10.5^(3x-5)
xlog75.2 = (3x-5)log10.5

Trying the next line... Suppose its the same as last time but the other way around?

Edit:
(xlog75.2=3x-5)/log10.5

 

[BruceW]

You got the last question right, with -0.75.

For the new question, you have got the use of logarithm correct. Now it is just a case of rearranging.

 

[lloydowen]

The new question I re-wrote the equation like so, xlog75.2 = 3xlog10.5-5log10.5 Is that wrong and/or a better method than the last way?

 

[Mark44]

The new question I re-wrote the equation like so, xlog75.2 = 3xlog10.5-5log10.5 Is that wrong and/or a better method than the last way?

You're not done yet. Add -3xlog(10.5) to both sides. Then on the left side, you'll have two terms with a factor of x. Factor the two terms into x(...other stuff) = <stuff on the right side>.

 

[cepheid]

You cannot put equals signs inside of expressions within parentheses. It makes no sense.

 

[lloydowen]

You cannot put equals signs inside of expressions within parentheses. It makes no sense.

Sorry, not quite sure what you mean here?

You're not done yet. Add -3xlog(10.5) to both sides. Then on the left side, you'll have two terms with a factor of x. Factor the two terms into x(...other stuff) = <stuff on the right side>.

Thanks for your help Mark, but I'm still unsure now :( Would it look like this..
(-3xlog(10.5)xlog75.2=-3xlog(10.5) 3x-5) / log10.5

Or ((-3xlog(10.5)xlog75.2=(3xlog(10.5))3xlog10.5-5log10.5) CONFUSED ! :(

 

[cepheid]

Sorry, not quite sure what you mean here?

I was talking about this:

(xlog75.2=3x-5)/log10.5

You have an equals sign inside the parentheses. It makes no sense.

 

[BruceW]

You have an equals sign inside the parentheses. It makes no sense.

haha, yeah, I think he means 'divide both sides by log10.5'

Edit: I don't know if that is allowable notation in maths, but you can see what he means by it.

 

[BruceW]

CONFUSED ! :(

Think it through yourself, you've done the hard bit already, you just needed to rearrange to get x as the subject of the formula.

 

[BruceW]

The new question I re-wrote the equation like so, xlog75.2 = 3xlog10.5-5log10.5 Is that wrong and/or a better method than the last way?

This bit is correct. Just think of the numbers as letters, you have something like: ax = 3bx - 5b, which is elementary algebra.

 

[lloydowen]

This bit is correct. Just think of the numbers as letters, you have something like: ax = 3bx - 5b, which is elementary algebra.

Ok... Thanks, I'll have another go now, My brain is fried haha!

 

[BruceW]

No worries, I know the feeling

 

[lloydowen]

I really don't see what I can do here, nothing seems to look right! I can see on your equation it would be (x=3bx-5b)/a but I just can't see how to apply that to my equation :(

 

[BruceW]

Your equation is xlog(75.2)=3xlog(10.5) - 5log(10.5) So you could also write it as ax=bx+c if you wanted. You can choose how to define your constants.

You don't even need to write it as algebraic letters. You could just write down the actual numbers straight away, then find what x must be.

 

[lloydowen]

So I should put the -5log(10.5) onto the left side, and cancel down? So I would be left with xlog(75.2)=3xlog(10.5) Or should I divide the right hand side by log(75.2) to get x on it's own? I don't think I've been taught this properly :/

 

[BruceW]

hmm, I think you should practise more at this kind of stuff then. Because it can come up a lot. Its called 'rearranging to get something as the subject of the formula', or 'collecting like terms', or could be called 'solving the equation'. Do you have a maths textbook? Looking through it again might help.

There are essentially a few rules you have to stick to when you're rearranging an equation. You can do things like 'multiply both sides by the same thing' and 'add something to both sides' I'm guessing you're familiar with these?

Using these rules, you cannot turn the equation xlog(75.2)=3xlog(10.5)-5log(10.5) into the equation xlog(75.2)=3xlog(10.5)

 

[lloydowen]

hmm, I think you should practise more at this kind of stuff then. Because it can come up a lot. Its called 'rearranging to get something as the subject of the formula', or 'collecting like terms', or could be called 'solving the equation'. Do you have a maths textbook? Looking through it again might help.

There are essentially a few rules you have to stick to when you're rearranging an equation. You can do things like 'multiply both sides by the same thing' and 'add something to both sides' I'm guessing you're familiar with these?

Using these rules, you cannot turn the equation xlog(75.2)=3xlog(10.5)-5log(10.5) into the equation xlog(75.2)=3xlog(10.5)

I am familiar with those yes, but where I am taught a tutor will go through a subject/topic and then "its upto you to learn it" but I try my best to learn things myself but it's really quite hard for me to understand... and I don't have a Maths textbook, only class notes, do you recommened a book I get?

I know what I have to do I just struggle to see how I do it, or struggle to find the next step? It's hard to explain really, but when I 'click' I can do the questions all the time, just having trouble with these ones :(

 

[Mark44]

I really don't see what I can do here, nothing seems to look right! I can see on your equation it would be (x=3bx-5b)/a but I just can't see how to apply that to my equation :(

Again, it makes no sense to write an equals in the middle of a parenthesized expression.

Here are the steps, starting from ax = 3bx−5b
ax - 3bx = -5b ; add -3bx to both sides
x(a - 3b) = -5b ; bring out common factor

x = -5b/(a - 3b) ; divide both sides by a - 3b

If you are trying to learn this without a textbook, that's not a good idea. I don't have any particular book to recommend, but there's a website that has tutorials in many different areas of mathematics, and this might be helpful - khanacademy.org.

 

[lloydowen]

Thank you for that! Tomorrow I'll get around to look that site and some online tutorials, I have tried them before but I've never really had much luck. Anyway good night from me, it's pretty late right now! :) Thanks for all your help guys !

 

[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.