Question about precedence rules

[solve]

Homework Statement

1-2+3=2:

a) 1-2=-1, -1+3=2

or

b) -2+3=1, 1+1=2

So you can attack this expression from either side and you still get 2.

My question is :

If I had to remove logs from lnP= 1/2ln(Q+1)- 3lnR+ 2

I'd either get P=[(Q+1)^1/2]/(R3* e^2) or P=[[(Q+1)^1/2]* e^2]/R^3 depending on which operation( taking the difference or adding) I'd choose to do first.

Which one is more correct and why does the answer depend on what(subtracting or addition) I do first? In case of 1-2+3=2 it doesn't.

Where is my logic going all illogical?

Thanks.

 

[Infinitum]

Here's how I interpreted the equation

$$ln(P) = \frac{1}{2}ln(Q+1) - 3ln(R) + 2 \\ ln(P) = ln(Q+1)^{1/2} - ln(R)^{3} + ln(e^2) \\ ln(P) = ln(\frac{{}e^2\sqrt{Q+1}}{R^3})$$

It doesn't matter in which order you take them, the terms with plus are always in multiplication, and terms with minus are divided.

 

[solve]

Here's how I interpreted the equation

$$ln(P) = \frac{1}{2}ln(Q+1) - 3ln(R) + 2 \\ ln(P) = ln(Q+1)^{1/2} - ln(R)^{3} + ln(e^2) \\ ln(P) = ln(\frac{{}e^2\sqrt{Q+1}}{R^3})$$

It doesn't matter in which order you take them, the terms with plus are always in multiplication, and terms with minus are divided.

Thanks for the reply, Infinitum. Your interpretation is correct. Let me do addition first, here:

lnP= 1/2ln(Q+1)- 3lnR+ 2

lnP= ln(Q+1)^1/2- (lnR^3+ lne^2)

lnP= ln[(Q+1)^1/2]/ln(R^3* e^2)

lnP= ln{[(Q+1)^1/2]/(R^3* e^2)}

P=[(Q+1)^1/2]/(R^3* e^2)

Please, show me where I am going wrong with this. Thanks.

 

[Infinitum]

lnP= ln(Q+1)^1/2 - lnR^3+ lne^2

lnP= ln[(Q+1)^1/2]/ln(R^3* e^2)

This step.

The mistake is that you did not take the minus common from the two before doing this.

$$ln(P) = ln(Q+1)^{1/2} - ln(R)^{3} + ln(e^2)$$

$$ln(P) = ln(Q+1)^{1/2} - (ln(R)^{3} - ln(e^2))$$

$$ln(P) = ln(Q+1)^{1/2} - ln(\frac{R^3}{e^2})$$

 

[solve]

This step.

The mistake is that you did not take the minus common from the two before doing this.

$$ln(P) = ln(Q+1)^{1/2} - ln(R)^{3} + ln(e^2)$$

$$ln(P) = ln(Q+1)^{1/2} - (ln(R)^{3} - ln(e^2))$$

$$ln(P) = ln(Q+1)^{1/2} - ln(\frac{R^3}{e^2})$$

Goodness. I just can't stop making such stupid mistakes. Thank you, Infinitum.

 

[Infinitum]

Goodness. I just can't stop making such stupid mistakes. Thank you, Infinitum.

You're welcome!

 

[solve]

You're welcome!

Wait, I think I have another question :D

Let's say I isolate lnR^3+ lne^2 with parenthesis so that I remember to do it first:

lnP= ln(Q+1)^1/2- (lnR^3+ lne^2)

Then I open the parenthesis:

lnP= ln(Q+1)^1/2- lnR^3- lne^2

it works out to the right answer:

P=[e^2*(Q+1)^1/2]/R^3. Fine.

But now I got 1-2+3=2

Lets say I do the same with it:

1-(2+3)=1-2-3=-4

2≠- 4.

Where did I stumble?

Thanks.

 

[Infinitum]

But now I got 1-2+3=?

Lets say I do the same with it:

1-(2+3)=1-2-3=-4

2≠- 4.

Where did I stumble?

Thanks.

The same mistake yet again

Your actual question is 1-2+3.

This becomes, 1-(2-3) ...golden rule, always, opening the bracket term should give you the original term.

 

[solve]

The same mistake yet again

Your actual question is 1-2+3.

This becomes, 1-(2-3) ...golden rule, always, opening the bracket term should give you the original term.

So, I guess, doing this wasn't entirely correct to start with:

"Let's say I isolate lnR^3+ lne^2 with parenthesis so that I remember to do it first:

lnP= ln(Q+1)^1/2- (lnR^3+ lne^2)"

I see. Parenthesizing for convenience without manipulating the signs to keep the original value intact is tres stupid.

Good. Nothing is too obvious for me. Many appreciations, Infinitum.

 

[solve]

Hi, All

I'd like to look at this situation once again.

Ok, so

1-2+3=
I can subtract 2 from 1 and, THEN add 3:
-1+3=2
or
I can add -2 and 3 first and then add 1:
1+1=2

Either way, I get the same answer and DON'T have to use parenthesis like 1-(2-3) if I want to add -2 and 3 first and then add 1. It's just not necessary.

But I ABSOLUTELY have to parenthesize if I want to add lnK and lne^KL before doing other operations inside {...}- parenthesis:

lnI= ln(2V)- { ln(KR+r)- lnK+ KL }

lnI= ln(2V)- { ln(KR+r)- (lnK- KL) }

I DIDN'T have to use parenthesis when I added negative 2 to 3 before adding 1 in 1-2+3 to get two, though, I could, of course, do 1-(2-3).

So why should I ABSOLUTELY use parenthesis in this particular case (log expression) and not in 1-2+3 if I want to start with adding the last two terms first?

Thanks.

 

[Infinitum]

Hi, All

I'd like to look at this situation once again.

Ok, so

1-2+3=
I can subtract 2 from 1 and, THEN add 3:
-1+3=2
or
I can add -2 and 3 first and then add 1:
1+1=2

Either way, I get the same answer and DON'T have to use parenthesis like 1-(2-3) if I want to add -2 and 3 first and then add 1. It's just not necessary.

But I ABSOLUTELY have to parenthesize if I want to add lnK and lne^KL before doing other operations inside {...}- parenthesis:

lnI= ln(2V)- { ln(KR+r)- lnK+ KL }

lnI= ln(2V)- { ln(KR+r)- (lnK- KL) }

I DIDN'T have to use parenthesis when I added negative 2 to 3 before adding 1 in 1-2+3 to get two, though, I could, of course, do 1-(2-3).

So why should I ABSOLUTELY use parenthesis in this particular case (log expression) and not in 1-2+3 if I want to start with adding the last two terms first?

Thanks.

No, you do -not- absolutely need to use parenthesis to simply even the log equation. Parenthesis is only an easier, understandable way of looking at it. If you are careful enough, you can avoid mistakes without parenthesis.

For,

$$lnI= ln(2V)- (ln(KR+r)- lnK+ lne^{KL})$$

You can simply write that as

$$lnI= ln(2V)- ( ln(KR+r) + ln\frac{e^{KL}}{K})$$

In loose terms, just remember, for logarithms, the terms with the negative sign go to denominator, and terms with a positive sign go to numerator. Or, see it this way as in the below example

$$-logK + logP$$

$$log\frac{1}{K} + logP$$

$$log \frac{P}{K}$$

 

[solve]

No, you do -not- absolutely need to use parenthesis to simply even the log equation. Parenthesis is only an easier, understandable way of looking at it. If you are careful enough, you can avoid mistakes without parenthesis.

For,

$$lnI= ln(2V)- (ln(KR+r)- lnK+ lne^{KL})$$

You can simply write that as

$$lnI= ln(2V)- ( ln(KR+r) + ln\frac{e^{KL}}{K})$$

See, to get ln(K/e^KL) you had to get { ln(KR+r)- lnK+ KL } to be { ln(KR+r)- (lnK- KL) }. In other words, you did have to parenthesize lnK+ KL. Otherwise, you'd just get K*e^KL.

 

[Villyer]

I can add -2 and 3 first and then add 1

Notice how you took that 2 as a negative 2 when you were showing the addition example.

When you did the log example, you removed the negative. You took both the second and the third term as being positive, and thus multiplied them. If you isolate the last two terms while still treating the second one negative, then it will end up in the denominator as it should.

 

[solve]

Notice how you took that 2 as a negative 2 when you were showing the addition example.

When you did the log example, you removed the negative. You took both the second and the third term as being positive, and thus multiplied them. If you isolate the last two terms while still treating the second one negative, then it will end up in the denominator as it should.

Noice! Also, can the explanation "Because addition and subtraction are commutative. Logarithms are not." be perfectly applicable here?

Thanks.

 

[Infinitum]

Noice! Also, can the explanation "Because addition and subtraction are commutative. Logarithms are not." be perfectly applicable here?

Thanks.

Nooo! Subtraction is not commutative.

$$a - b \neq b - a$$

For all a,b belonging to ℝ.

Logarithm addition is commutative.

$$log(m) + log(n) = log(n) + log(m)$$

 

[solve]

Nooo! Subtraction is not commutative.

$$a - b \neq b - a$$

For all a,b belonging to ℝ.

Logarithm addition is commutative.

$$log(m) + log(n) = log(n) + log(m)$$

Thanks for catching that.