Logarithm question: I can't figure out how (log5)^2+(log2)^2+2log2log5 equals 1

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Helenam

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Just started doing logs again and got a bit stuck with this.
(log5)^2+(log2)^2+2log2log5
I can’t figure out how this = 1
 
Just for curiosities sake. Is there any other way to solve this using the basic laws of logs?
 
Helenam said:
Just for curiosities sake. Is there any other way to solve this using the basic laws of logs?
Click to expand...
You have the following:

[imath]\qquad \log^2(5) + \log^2(2) + 2\log(5)\log(2)[/imath]

As pointed out earlier, this is a perfect-square trinomial:

[imath]\qquad \left(\log(5) + \log(2)\right)^2[/imath]

Now apply the log rule that allows conversion between the log of a product and a sum of logs. Then notice that this log has a base of [imath]10[/imath].

Where does this lead?
 
Helenam said:
Is there any other way to solve this using the basic laws of logs?
Click to expand...
After using the "reduction" suggested by response #2 - further simplification can be achieved using basic laws of logs

→ logb(a) + logb(c) = logb(a*c) ................... and continue...........
 
Helenam said:
Just started doing logs again and got a bit stuck with this.
(log5)^2+(log2)^2+2log2log5
I can’t figure out how this = 1
Click to expand...
Helenam said:
Just for curiosities sake. Is there any other way to solve this using the basic laws of logs?
Click to expand...
To be honest I looked at [imath](\log(2))^2+(\log(5))^2+2\log(2)\log(5)[/imath] and saw at once it equals [imath]\left(\log(10)\right)^2[/imath]
But it took sometime to realize why the answer given is [imath]1[/imath].
I had no idea that anyone in professional mathematics still uses [imath]\log(x)[/imath] for anything other than the logarithm.
You question should be written as [imath](\log_{10}(2))^2+(\log_{10}(5))^2+2\log_{10}(2)\log_{10}(5)[/imath]
Then [imath]\left(\log_{10}(10)\right)^2=1[/imath]

[imath][/imath]
 
pka said:
I had no idea that anyone in professional mathematics still uses log⁡(x)\log(x)log(x) for anything other than the logarithm.
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Yes, professional mathematicians interpret "log" as base e, but most students are used to log as base 10 and ln as base e, as depicted on most scientific calculators keys.
 
Physicists (at least the one's I've hung around) still use the expression ln(x). I almost never use any other base, in fact. Honestly, I have to think about it when I see someone use log(x) and not mean [imath]\log_{10}(x)[/imath]

-Dan
 
My question is why does professional mathematicians and students have different definitions for the base for log? I thought that professional mathematicians are the ones who write the text books!
 
Steven G said:
My question is why does professional mathematicians and students have different definitions for the base for log? I thought that professional mathematicians are the ones who write the text books!
Click to expand...
I'm "old school." Way back when (and maybe there was even a different convention before this) I was using textbooks written in the early 80's to learn Math. That was when ln was ln and log was [imath]\log_{10}[/imath]. At some point (the late 90's?) that changed and log became [imath]\log_e[/imath].

Kind of like that I still use atn(x) instead of [imath]tan^{-1}(x)[/imath] because it's sooo much easier to write!

-Dan
 
topsquark said:
Physicists (at least the one's I've hung around) still use the expression ln(x). I almost never use any other base, in fact. Honestly, I have to think about it when I see someone use log(x) and not mean [imath]\log_{10}(x)[/imath]

-Dan
Click to expand...
It's bad enough that you're a Physicist, buy now I learn that you hung out with Physicists-Oh My.

Is 200*50=100 an acceptable result to you? After all, it's only off by a power of 10.
Why do Physicists not care about being off by powers of 10?
Isn't that why the space shuttle went down?
 
topsquark said:
I'm "old school." Way back when (and maybe there was even a different convention before this) I was using textbooks written in the early 80's to learn Math. That was when ln was ln and log was [imath]\log_{10}[/imath]. At some point (the late 90's?) that changed and log became [imath]\log_e[/imath].

Kind of like that I still use atn(x) instead of [imath]tan^{-1}(x)[/imath] because it's sooo much easier to write!

-Dan
Click to expand...
I sometimes write arctan but never have written atan. I have seen and understand atan.
 
Steven G said:
It's bad enough that you're a Physicist, buy now I learn that you hung out with Physicists-Oh My.

Is 200*50=100 an acceptable result to you? After all, it's only off by a power of 10.
Why do Physicists not care about being off by powers of 10?
Isn't that why the space shuttle went down?
Click to expand...
No, 200*50 = 100 is ridiculous.

Spherical horses, though, those are real!

-Dan
 
topsquark said:
No, 200*50 = 100 is ridiculous.

Spherical horses, though, those are real!

-Dan
Click to expand...
I once was having trouble getting the correct answer to a physics problem so I went to my physics professor. He looked at my work and said it was correct. I pointed out that the books answers was different. He looked at the answer in the book, saw that my answer differed from the books answer by a power of ten and then he through me out of his office saying that they were the same answers. This is one of a few reasons why I decided to study math rather than physics.
I am pleased to to see that that you agree that those extra zero's at the end of an answer matters.
 
You could solve the problem as follows:

Let x=log5 and y = log2.
Then the problem becomes x^2 + y^2 + 2xy = (x+y)^2
Now x+y = log 5 + log2 = log(5*2) = log 10 = 1.
So, x^2 + y^2 + 2xy = (x+y)^2= (1)^2 = 1
 
Yes thanks. I’ve got it now. I appear to have started an interesting conversation. My textbooks use log to mean log base 10 and ln for log base e. Unfortunately, I am a long way from being a professional mathematician.
 
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