Solve Simple Logarithms: log91/27

[matadorqk]

Homework Statement

#8: Simplify: log91/27

Homework Equations

y=logb(x)
x=b^y

The Attempt at a Solution

I have personally not tried logarithms before, but I think the problem is really easy and I am probably just over-thinking it?

I think that, using the formula's below, 9^a=1/27
So, a=1/243, so my simplified answer should be 9^1/243=1/27..
Can I simplify fractions then, so that 9^1/9=1?
I have personally no exact knowledge on what I am doing, but can someone give me some hints or basic concepts on logarithms? Thanks!

**Edit:
Ok, I might be wrong above, I found some more laws, and since loga (x/y)= logax - loga y,
I get:
log 9 (1) - log 9 (27)

since loga 1 = 0
I know have:
- log 9 (27)
Err.. now what?
I have no idea how to plot it into my calculator...
But from my formulas (first two) I get that 9^2/3=27.. but does simplifying mean turning into exponential form? :S

 

[genneth]

27 = 3^3 = 9^(3/2) ==> log_9(27) == 3/2

 

[matadorqk]

27 = 3^3 = 9^(3/2) ==> log_9(27) == 3/2

Huh? I'm pretty sure its 2/3... maybe I am wrong but could you further explain yourself?

 

[rocomath]

log(base9) (1/27) = x

9^x = (1/27)

x = log(1/27)/log(9)

 

[matadorqk]

log(base9) (1/27) = x

9^x = (1/27)

x = log(1/27)/log(9)

So log (1/27) / log (9)

log (1/27) = - log 27 / log 9
Can I simplify to - log 3 / log (1)
So - log 3 / ..0?

Hmm, so I can't simplify or my answer is undefined...

Is my answer -log 27 / log 9..
Im really lost here.

How about this:
Since log (base a) (a) = 1, does log (base a) (3a) = 3? If so, log (base 9) (27) = 3?

 

[rocomath]

log(base9) (1/27) = x

9^x = (1/27) in exponential form ... since it said to simplify, i solved sorry

type in your calculator log(1/27)/log(9) and plug into x, 9^(-1.5) = 1/27

 

[matadorqk]

log(base9) (1/27) = x

9^x = (1/27) in exponential form ... since it said to simplify, i solved sorry

No problem. So, 9^x= 1/27 is simplified?..

*bangs head against wall* ( I knew it wasn't that complicated, :P)

 

[rocomath]

so the general rule, and with the relevant equations you gave should have been enough information for you to solve it

anyways, the way i learned to convert/simplify/solve logs is by how you read logs

log(base) (y) = x

the base must be raised to some power x, to get y

logs are just inverses, one-to-one functions

 

[matadorqk]

so the general rule, and with the relevant equations you gave should have been enough information for you to solve it

anyways, the way i learned to convert/simplify/solve logs is by how you read logs

log(base) (y) = x

the base must be raised to some power x, to get y

logs are just inverses, one-to-one functions

Thanks a lot for your help, ill go on to the next log problem :) hehe, I guess it wasn't so complicated, I just really didn't know what they ment by "simplify" but now I will assume that simplifying a log solely involves getting rid of the 'log'. Again, thanks.

 

[rocomath]

Thanks a lot for your help, ill go on to the next log problem :) hehe, I guess it wasn't so complicated, I just really didn't know what they ment by "simplify" but now I will assume that simplifying a log solely involves getting rid of the 'log'. Again, thanks.

i just reviewed logs Monday so it's fresh on my mind!

some can be a real b tho :) have fun

 

[Feldoh]

Actually simply means simplify until you cannot simplify anymore. If you can reach a solution that is how far you need to go, like in this case.

Here's how I did it (since you already have the answer :-/):

log_9(\frac{1}{27}) = \frac{log_{3}\frac{1}{27}}{log_{3}9}

\frac{log_{3}\frac{1}{27}}{log_{3}9} = \frac{log_{3}3^{-3}}{log_{3}3^2}

\frac{log_{3}3^{-3}}{log_{3}3^2} = \frac{-3}{2}

 

[learningphysics]

Huh? I'm pretty sure its 2/3... maybe I am wrong but could you further explain yourself?

27 = 9 * 3 = 9 * 9^{1/2} = 9^{1 + 1/2} = 9^{3/2}

1/27 = 27^{-1} = (9^{3/2})^{-1} = 9^{-3/2}

 

[matadorqk]

Actually simply means simplify until you cannot simplify anymore. If you can reach a solution that is how far you need to go, like in this case.

Here's how I did it (since you already have the answer :-/):

log_9(\frac{1}{27}) = \frac{log_{3}\frac{1}{27}}{log_{3}9}

\frac{log_{3}\frac{1}{27}}{log_{3}9} = \frac{log_{3}3^{-3}}{log_{3}3^2}

\frac{log_{3}3^{-3}}{log_{3}3^2} = \frac{-3}{2}

Wow... Thanks a lot for that! That explains a lot :P

 

[matadorqk]

log_9(\frac{1}{27}) = \frac{log_{3}\frac{1}{27}}{log_{3}9}

\frac{log_{3}\frac{1}{27}}{log_{3}9} = \frac{log_{3}3^{-3}}{log_{3}3^2}

\frac{log_{3}3^{-3}}{log_{3}3^2} = \frac{-3}{2}

Umm, may I ask how you get from log_9(\frac{1}{27}) = \frac{log_{3}\frac{1}{27}}{log_{3}9}

Is there a formula?
Like...: log(base a) (b) = log (base (square root of a)) (b) / log (base (square root of a)


You kind of make sense, but I want to know how to implement it in any problem..
I hope you don't mind explaining after you solved it...

 

[Feldoh]

Umm, may I ask how you get from log_9(\frac{1}{27}) = \frac{log_{3}\frac{1}{27}}{log_{3}9}

Is there a formula?
Like...: log(base a) (b) = log (base (square root of a)) (b) / log (base (square root of a)You kind of make sense, but I want to know how to implement it in any problem..
I hope you don't mind explaining after you solved it...

Change of base is what it is, most people don't realize you can change to any base and not just base 10.

log_{a}b = \frac {log_{n}b}{log_{n}a} where n is an integer and is the new base.

 

[matadorqk]

Change of base is what it is, most people don't realize you can change to any base and not just base 10.

log_{a}b = \frac {log_{n}b}{log_{n}a} where n is an integer and is the new base.

Umm, but leaving n as an integer doesn't that give room for n=any integer while a and b don't change? I am going to analyze this and ill get back to you.

 

[rocomath]

change-of-base formula is ... log(b)M=log(a)M/log(a)b where log(a) = log(base10)

let's start off with the Proof

Proof, let x = log(b)M ... in exponential form b^x = M

1) b^x = M

2) take the log(a) of both sides ... log(a)b^x = log(a)M

3) solve for x ... x = log(a)M/log(a)b

4) if x = log(b)M ... (step 3) log(b)M = log(a)M/log(a)b

... log(a) can be anything, it doesn't have to be log(base10)

 

[matadorqk]

change-of-base formula is ... log(b)M=log(a)M/log(a)b where log(a) = log(base10)

let's start off with the Proof

Proof, let x = log(b)M ... in exponential form b^x = M

1) b^x = M

2) take the log(a) of both sides ... log(a)b^x = log(a)M

3) solve for x ... x = log(a)M/log(a)b

4) if x = log(b)M ... (step 3) log(b)M = log(a)M/log(a)b

... log(a) can be anything, it doesn't have to be log(base10)

so... if my answer is log(3) 3^-3 / log (3) 3^2 = -3/2...
Can it also be log (5) 3^-3 / log (5) 3^2 = -3/2.. ?

Or does the new base have to be related to the original base? This is what has me confused.. How do you define the new base, and what effect does this have on the other terms of the equation?

This would be because even though even though the numbers may be larger/smaller, the ratio is the same? I hope I am right this time, I think I am getting this logs thing :)

 

[rocomath]

i think so ... lol, but I'm no expert.

 

[matadorqk]

Thanks, ill re-check with my teacher tommorow. Anyways, I got stuck again.

Homework Statement

Convert \frac{8}{\log_{5}9} to an expression of the form a\log_{3}b
Ok that's the problem.

Homework Equations

#1: y=\log_{a}(x)
#2: x=a^y

The Attempt at a Solution

**Im working on it as you read :) bare with me and I'll give you my attempt hehe

 

[rocomath]

hmm ... let me evaluate, i see that 8 can be re-written as 2^3 & 9 as 3^2 ... still thinking

... kicking my ass! hehe

is this correctly typed in?

 

[learningphysics]

Convert: \log_{5}9} to log base 3, using the method discussed in this thread...

 

[leright]

meh, I just figure the exponent must be negative if it is to produce a fractional result. So, then the problem just reduces to a matter of manipulating 9 to get 27. taking the square root of 9 yields 3 and cubing 3 yields 27. So, the exponent would need to be -2/3

 

[matadorqk]

Convert: \log_{5}9} to log base 3, using the method discussed in this thread...

Great!
So, 2^3 over: \frac{\log_{3}3^2}{\log_{3}5}?

meh, I just figure the exponent must be negative if it is to produce a fractional result. So, then the problem just reduces to a matter of manipulating 9 to get 27. taking the square root of 9 yields 3 and cubing 3 yields 27. So, the exponent would need to be -2/3

As much as that seems simpler and faster, if I don't learn the method to actually do it for any problem instead of solving 1, I'll have problems with similar probs. And um, I think we solved that one already, its actually -3/2

 

[learningphysics]

Great!
So, 2^3 over: \frac{\log_{3}3^2}{\log_{3}5}?

Sorry, yes. 8 over that. And that's it. So I get 4log_3{5}

 

[matadorqk]

Sorry, yes. 8 over that. And that's it. So I get 4log_3{5}

Err.. big jump there :P, What do I do when I have a.. three layer fraction?
..nvm.. Lemme figure this out

 

[learningphysics]

Err.. big jump there :P, What do I do when I have a.. three layer fraction?
say a/b/c= ca/b?

\frac{a}{(\frac{b}{c})} = \frac{ac}{b}

but: \frac{(\frac{a}{b})}{c}=\frac{a}{bc}

 

[matadorqk]

\frac{a}{(\frac{b}{c})} = \frac{ac}{b}

but: \frac{(\frac{a}{b})}{c}=\frac{a}{bc}

I used this formula and the one above,
\log_{a}x^y=y\log_{a}x

\frac{8(\log_{3}5)}{2\log_{3}3}

Right?

But then I get
4\log_{3}\frac{5}{3}

 

[learningphysics]

I used this formula and the one above,
\log_{a}x^y=y\log_{a}x

\frac{8(\log_{3}5)}{2\log_{3}3}

Right?

That's right.

But then I get
4\log_{3}\frac{5}{3}

this is wrong. (log_a{x})/(log_a{y}) is not equal to log_a{x/y}. what is log_3{3}?

 

[matadorqk]

That's right.



this is wrong. (log_a{x})/(log_a{y}) is not equal to log_a{x/y}. what is log_3{3}?

Ugh right! It equals 1!

So there we have it. \frac{8}{2}(log_{3}5) So: 4\log_{3}5 Yay!

Improvement!

 

[matadorqk]

Thanks evrn!

Homework Statement

Express as a single logarithm 2lnx + ln (x-1) - ln (x-2)

Homework Equations

None used (Basic Algebra)

The Attempt at a Solution

So, let's substitute ln for a, to make it simpler
2ax +a (x-1) - a (x-2)
2ax +ax - a - ax +2a
2ax - a +2a
2ax+a
2lnx+ln
ln (2x+1)

I think that's correct, I'd appreciate some re-check just in case though.

That's the last one, thanks everyone for the help!

 

[learningphysics]

Homework Statement

Express as a single logarithm 2lnx + ln (x-1) - ln (x-2)

Homework Equations

None used (Basic Algebra)

The Attempt at a Solution

So, let's substitute ln for a, to make it simpler
2ax +a (x-1) - a (x-2)
2ax +ax - a - ax +2a
2ax - a +2a
2ax+a
2lnx+ln
ln (2x+1)

I think that's correct, I'd appreciate some re-check just in case though.

That's the last one, thanks everyone for the help!

You can't substitute for ln... ln is not a variable. it's a function... You could substitute for something like ln(a) which is a value, or ln(x)... but not just ln...

 

[rocomath]

2lnx = ln(x^2)

ln(x-1) - ln(x-2) = ln[(x-1)/(x-2)]

ln(x^2) + ln[(x-1)/(x-2)] = ln[(x^3-x^2)/(x-2)]

 

[matadorqk]

You can't substitute for ln... ln is not a variable. it's a function... You could substitute for something like ln(a) which is a value, or ln(x)... but not just ln...

Uh oh.. Ok let's restart again then.

So ln= log_{e}x

Ok, ill go from there and get back to you

**Or take roco's idea, a head start :)

 

[rocomath]

**Or take roco's idea, a head start :)

nooo!

 

[matadorqk]

nooo!

... I won't just copy it and not learn no worries.
A couple questions:

Is 2lnx=lnx^2 a formula or 'standard'?
Ohh wait, do you get that from the log formulas? That explains alot!
So you are using \log_{a}x^y=y\log_{a}x

And then you use \log_{a}(\frac{x}{y})=\log_{a}x-\log_{a}y..

So that's how you do your first two steps. Then you insert lnx^2 into the ln (x-1) in a reverse distribution. I think I got that correctly, but is that as much as I can simplify?

Would \frac{\ln(x^3-x^2)}{x-2} be the final answer? I stared at it for a while haha, can't find a way to simplify more.

 

[learningphysics]

... I won't just copy it and not learn no worries.
A couple questions:

Is 2lnx=lnx^2 a formula or 'standard'?
Ohh wait, do you get that from the log formulas? That explains alot!
So you are using \log_{a}x^y=y\log_{a}x

And then you use \log_{a}(\frac{x}{y})=\log_{a}x-\log_{a}y..

So that's how you do your first two steps. Then you insert lnx^2 into the ln (x-1) in a reverse distribution. I think I got that correctly, but is that as much as I can simplify?

Would \frac{\ln(x^3-x^2)}{x-2} be the final answer? I stared at it for a while haha, can't find a way to simplify more.

Careful. the entire fraction should be inside the ln...

 

[matadorqk]

Careful. the entire fraction should be inside the ln...

Right!

ln(\frac{x^3-x^2}{x-2})

 

[learningphysics]

Right!

ln(\frac{x^3-x^2}{x-2})

That's the answer. I don't think you can simplify any more.

 

[matadorqk]

Well, not to take the risk, there's one more problem I did without consulting, so I rather check with you to see if I did it correct. This one I feel a bit more confident about, let's see how that goes:
1. Homework Statement

Solve for x: \log_{3}x + \log_{3}(x-2)=1

2. Homework Equations

#1\log_{a}x+\log_{a}y=\log_{a}xy
#2y=\log_{b}(x)
#3x=b^y
3. The Attempt at a Solution

Ok so let's solve using the formula #1:

So:
\log_{3}(x)(x-2)=1
\log_{3}(x^2-2x)=1
To make this simpler, let's make x^2-2x=w
log_{3}(w)=1
So use formula #2/#3 to get:
w=3^1

x^2-2x=3

Now I kind of got confused on what to do. I seriously don't know why I can't algebraically solve it, by trial and error I got x=3.. but perhaps its the excessive coke or sometihng that's blocking me from algebraically solving this. Help?

 

[rocomath]

have you taken Chemistry II? it's heavily used for Acids/Bases and Reaction Rates, easier than this tho.

 

[matadorqk]

have you taken Chemistry II? it's heavily used for Acids/Bases and Reaction Rates, easier than this tho.

Nope.. I only took General Chemistry a year ago (10th)..

 

[learningphysics]

Have you learned how to solve quadratic equations?

 

[rocomath]

you're already at the 2nd to the last step

x^2 - 2x - 3 = 0

now solve for x

 

[matadorqk]

Have you learned how to solve quadratic equations?

OHH OK, I think I can solve that!

X=3 or x=-1

as (x-3)(x+1)=0

 

[learningphysics]

OHH OK, I think I can solve that!

X=3 or x=-1

as (x-3)(x+1)=0

Yes, but x=-1 is an extraneous root... so your answer is just x = 3.

 

[Feldoh]

To clarify it's extraneous because
\log_{3}(-1) + \log_{3}((-1)-2)=1

However we encounter a problem here since

log_{a}b = c, "b" must be >0

Clearly in this problem it's not and doesn't work.

 

[HallsofIvy]

To clarify it's extraneous because
\log_{3}(-1) + \log_{3}((-1)-2)=1

Well, no, it's not. log₃(-1) and log₃(-3) are undefined and have no value- their sum is NOT equal to 1, it doesn't equal anything.

However we encounter a problem here since

log_{a}b = c, "b" must be >0

Clearly in this problem it's not and doesn't work.

Yes, that's the reason -1 is "extraneous". It satisfies the equation x²- 2x= 3 but not log₃(x)+ log₃(x-1)= 1.

 

[Feldoh]

Well, no, it's not. log₃(-1) and log₃(-3) are undefined and have no value- their sum is NOT equal to 1, it doesn't equal anything.


Yes, that's the reason -1 is "extraneous". It satisfies the equation x²- 2x= 3 but not log₃(x)+ log₃(x-1)= 1.

\log_{3}(-1) + \log_{3}((-1)-2)=1 was just to show an easy way to check -- plug the numbers into the equation. Doesn't match the domain of a log so it doesn't exist...