Comment by RiderOfGiraffes | Hacker News Reader

Comment by RiderOfGiraffes | Hacker News Reader

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RiderOfGiraffes

But you still don't have b as the subject of the equation. You need to get to "b=..." If you don't know how to do that, then at least tell us what you've tried.

Can you make x the subject of this equation: log(x)+3=2t ? In other words, if the equation log(x)+3=2t is always satisfied, what does x equal?

And I will look at the partial fractions, but I'd like to concentrate on this first.

Actually, what is your definition of a logarithm?

The last time we had a conversation I asked you some questions to assess your level of knowledge and you never answered them:

http://news.ycombinator.com/item?id=1107996

It seems to me looking over other times you've asked questions that you claim you want to get into math, and you probably think you're doing lots of stuff, but in reality I've seen no evidence of you actually trying things.

Gaining skills is a matter of spending the time. In math in particular, if you haven't spent ages getting loads of wrong answers then you won't have any real intuition as to what to do. You need loads of wrong answers, from which you can start to identify "good wrong answers". From those good wrong answers you can build an intuition as to what might work, and you can start to find excellent wrong answers.

Then the moment comes when you see how to convert an excellent wrong answer into a right answer.

With practice this becomes second nature, and after a time you wonder why you ever found it difficult. But until you've put in the hours of practice, it doesn't happen.

People think doing math is a matter of somehow finding the right process and then following it. "Tell me what to do" is the perpetual cry.

You need to try stuff. Try re-arranging the equations repeatedly into different forms until, maybe, one looks familiar. Don't just stare at stuff hoping inspiration will strike, and don't think there's a magic formula for finding the right next step.

That's not how it works.

So tell me, how many different ways can you rearrange the equation: log(x)+3=2t ?

i completely agree with you sir and i really appologise for not responding to this thread http://news.ycombinator.com/item?id=1107996 to be honest with you i really don't know where i stand .what's my level of competence. i recently gained interest in math and physics because i regret not taking these two subjects seriously back in college.now i work for a bank which pay's peanuts.i don't intend to get a job which helps me build my math/physics skills,i just wanna learn it because i have a burning desire to learn it.it's as simple as that. yes.math requires hours and days of practice which i'am religiously doing but it's not enough for me as it's very evident by the level of knowledge/intellect that i possess. with that being said the number of ways arranging the equation log(x)+3=2t hmm.. log(x)=2t-3 am i right??hope i'am not bullshitting you.

by the way the only definition of logarithm that i know is 'log is inverse of exponents'

So what does that mean?

OK, here are some terms to simplify. In each case, what is a simpler way to write the expression:

  x^3 * x^5

  (t^3)^2

  u^(3^2)

Expand the following:

  (g+2)^2

  (z-3)^3

Now, if x=10^y, what does y equal?

x^3+5=x^8

t^3 * t^3 = t^6

i guess it's u^6

g^2+4g+4

not sure abt (z-3)^3

expanding (z-3)^3

could it be: (z-3).(z+3)^2 ??

=>(z-3) (z^2+3z+9)

OK, in the first one you need brackets : x^3+5 is not usually understood to be the same as x^(3+5), but the answer is right, and the process is right. The others are right too.

So (g+2)^2 is equal to g^2+4g+4, but it's also equal to (g+2).(g+2) [where I've used the dot for multiplication]

Does that help you for (z-3)^3 ??

now i have the solution for y in x=10^y

here it is:x=10^y

subtract 10^y from both the sides

x-10^y=0

now subtract x from both the sides

-10^y = -x

divide by -1

-10^y/-1 = -x/-1

= 10^y = x

take log base 10 both sides

log10^y = logx

= ylog10=logx

y = logx/log10

Ah, very good. You've gone slightly the long way around, but you've got the right answer.

The first part you go from x=10^y to -10^y=-x to 10^y=x. Is it not clear that if x=10^y then 10^y=x? The point is that any time A=B, then B=A.

Next, you have ended with y=log(x)/log(10). If the logarithms are base 10, what is log(10)? Can you use that to make your result simpler?

Next question: if t=log(u) (where the logarithm is base 10) then what does u equal?

i'll be back in 45 min

i'am not able to solve t=log(u).not even a single step sorry

i'am assuming this would end up in some number called 'e' which i came accross somewhere in the past.please help..i'am not sure what that number means and how to get it

OK, I really don't know how to help you now using this medium. You claim to have watching the Khan Academy video, but this question is answered in that.

The simple rules are these:

    log(a.b) = log(a) + log(b)

  hence

    log(a^4) = log(a.a.a.a) = log(a)+log(a)+log(a)+log(a) = 4.log(a)

  Therefore

    if  a = b^z
    then log(a) = log(b^z)
    so   log(a) = z.log(b)

This is true for all bases. Then we have

    log_b(b) = 1  :  Log (base b) of b is 1.

    Hence log_b(b^2) = 2
          log_b(b^3) = 3

    and so on.

In all this the basic rule is this:

    if   x = b^y
    then log_b(x) = log_b(b^y)
                  = y.log_b(b)
                  = y

So if the base is 10, then we have:

    If  a = 10^b
    then log(a) = log(10^b)
                = b.log(10)
                = b.1
                = b

So now, start with the equation t=log(u), assuming the log is base 10, and use the above rules to change it around. When you do that, see if you can get to saying u equals something.

Finally, you need to study sections 1, 2 and 3 of this page:

http://en.wikipedia.org/wiki/Logarithm

What's on that page will also help you to understand the questions.

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