Urgent! Need help understanding Logarithm Question!

[SirChadimus]

Also, another similar question that has 10^X but division IS solved by using the log numbers.

log 10^-4 / 10^-12

= -4-(-12) = 8


So why do we use the exponents(logs) for this, but not for addition!??

 

[SirChadimus]

Hello,

I'm having some trouble understanding a specific concept of logarithms, preparing for a quiz tommorow.

I understand the following laws:

log ab = log a + log b

and

log a + log b = log ab (antilog)

On the "practice test", we are posed with two questions like this:

1) log (3 * 3) = X

And I would solve by...

log3 + log3 = .477 + .477 = .954. The Correct Answer.

But then, I get similarly:

2) log (10^6+10^6) = X

And I have no idea how they got the answer they did, 6.301

They show the work, but it makes no sense.

Based on the law of the anti-log, it should be the log of 10^6 X 10^6, which would be 6x6, wouldn't it? Since 10 is the base and 6 is the log?

Instead, they provide the following work...

10^6 = 1 x 10^6 = 1,000,000

10^6 + 10^6 = 2 x 10^ 6 <--------------- Why did they do this!? Doesn't this completly go against the anti-log law!???

log (10^6 + 10^6) = log (2 x 10^6) <------- ???

log (2 x 10^6) = log 2 + log 10^6 = 0.301 + 6.000 = 6.301



Also, another similar question that has 10^X but division IS solved by using the log numbers.

log 10^-4 / 10^-12

= -4-(-12) = 8


So why do we use the exponents(logs) for this, but not for addition!??If anyone could help me understand this, I would be grateful!

 

[tkhunny]

Let's look at this VERY closely.

log ab = log a + log b

The left-hand-side has two values, "a" and "b" which are MULTIPLIED INSIDE the argument of the logarithm.
The right-hand-side has two values, "log(a)" and "log(b)" which are ADDED OUTSIDE the argument of any logarithm.

The Log Rule
Multiplication Inside leads to Addition Outside.

The AntiLog Rule
Addition Outside leads to Multiplication Inside

Now, let's look at your problem statement:

\(\displaystyle log(10^{6} + 10^{6})\)

This expression has ADDITION INSIDE the logarithm argument. We have not defined a rule for that. We need MULTIPLICATION INSIDE.

In order to get something useful, we simply use addition.

x + x = 2x
Frog + Frog = 2*Frog
10^6 + 10^6 = 2*(10^6)

Really, it's just addition.

\(\displaystyle log(10^{6} + 10^{6}) = log(2\cdot 10^{6})\)

NOW we use the logarithm rules.

\(\displaystyle log(10^{6} + 10^{6}) = log(2\cdot 10^{6}) = log(2) + log(10^{6}) = log(2) + 6\)

That last step should be obvious because we are using Base 10 Logarithms.

Are we starting to see it?

 

[SirChadimus]

Wow thanks for the reply,

However, the only thing I don't understand is why we use 2? Forgive my ignorance, It's been a while doing this kind of math for me.

Why does log (10^6+10^6) = log (2+10^6).

If since its an "Inside equation" and we need to use the product law, wouldn't it be 10^6 x 10^6?

 

[JeffM]

Hello,

I'm having some trouble understanding a specific concept of logarithms, preparing for a quiz tommorow.

I understand the following laws:

log ab = log a + log b

and

log a + log b = log ab (antilog) What does this have to do with anti-logs. If x = log(a), y = log(b), and z = log(ab), then x+ y = z,
which means that z = x + y. Whatever you believe to be the "law of anti-logs," you had best review it because you said the SAME law of logs in two different but equivalent ways and never referenced anti-logs.

On the "practice test", we are posed with two questions like this:

1) log (3 * 3) = X This is about 3 TIMES 3

And I would solve by...

log3 + log3 = .477 + .477 = .954. The Correct Answer.

But then, I get similarly:

2) log (10^6+10^6) = X This is about 10^6 PLUS 10^6. It is not true that log(a + b) = log(ab), which is in fact = log(a * b).

And I have no idea how they got the answer they did, 6.301

They show the work, but it makes no sense.

Based on the law of the anti-log, it should be the log of 10^6 X 10^6, which would be 6x6, wouldn't it? Since 10 is the base and 6 is the log? Wow. You just said that 1,000,000 + 1,000,000 = 1,000,000 * 1,000,000 and that the log(1,000,000) = log(6).

Instead, they provide the following work...

10^6 = 1 x 10^6 = 1,000,000

10^6 + 10^6 = 2 x 10^ 6 <--------------- Why did they do this!? Doesn't this completly go against the anti-log law!??? Because 1,000,000 + 1,000,000 = 2,000,000.

log (10^6 + 10^6) = log (2 x 10^6) <------- ???

log (2 x 10^6) = log 2 + log 10^6 = 0.301 + 6.000 = 6.301



Also, another similar question that has 10^X but division IS solved by using the log numbers.

log 10^-4 / 10^-12

= -4-(-12) = 8


So why do we use the exponents(logs) for this, but not for addition!??

log(a + b) does not have some nice clean equivalent expression in logarithms. Logarithms were originally invented to help do multiplication, division, exponentiation, and extracting roots before calculators and computers. Sums of logarithms represent the logarithm of a product, not a sum. I am not sure where you have got mixed up, but I suggest re-reading your text very carefully.

If anyone could help me understand this, I would be grateful!

.

 

[tkhunny]

I tried to convince you with two examples. Let's give it another go.

x + x = 2x -- It's just addition. Nothing to do with logarithms.

Frog + Frog = 2*Frog -- It's just addition. Nothing to do with logarithms.

\(\displaystyle \sqrt{2} + \sqrt{2} = 2\cdot\sqrt{2}\) -- It's just addition. Nothing to do with logarithms.

You seem to want to change addition to multiplication by MAGIC. Don't do that. Use the logarithm rules.

Let's try those examples again.

log(x + x) = log(2x) -- It's just addition. Nothing to do with logarithms.

log(Frog + Frog) = log(2*Frog) -- It's just addition. Nothing to do with logarithms.

\(\displaystyle log(\sqrt{2} + \sqrt{2}) = log(2\cdot\sqrt{2})\) -- It's just addition. Nothing to do with logarithms.

Closer?

 

[SirChadimus]

I tried to convince you with two examples. Let's give it another go.

x + x = 2x -- It's just addition. Nothing to do with logarithms.

Frog + Frog = 2*Frog -- It's just addition. Nothing to do with logarithms.

\(\displaystyle \sqrt{2} + \sqrt{2} = 2\cdot\sqrt{2}\) -- It's just addition. Nothing to do with logarithms.

You seem to want to change addition to multiplication by MAGIC. Don't do that. Use the logarithm rules.

Let's try those examples again.

log(x + x) = log(2x) -- It's just addition. Nothing to do with logarithms.

log(Frog + Frog) = log(2*Frog) -- It's just addition. Nothing to do with logarithms.

\(\displaystyle log(\sqrt{2} + \sqrt{2}) = log(2\cdot\sqrt{2})\) -- It's just addition. Nothing to do with logarithms.

Closer?

Thank you for your help, I think I understand now. My readings were pretty unclear about type of problem, so when it was thrown at me in the practice exam, I didn' tunderstand how to proceed. Thank you!

 

[HallsofIvy]

Why does log (10^6+10^6) = log (2+10^6).

It doesn't! What makes you think it does? What is true is that log(10^6+ 10^6)= log(2 X 10^6)= log(2)+ log(10^6= log(2)+ 10^6, assuming your logarithm is base 10. That is NOT what you wrote.

 

[tkhunny]

Missed that. Assumed it was a typo, I guess.