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# Check a logarithm answer? watch

1. No answers in my book grrrr. Can I get a second opinion:

Find X

5 x 2^x = 10^x

I make x to be log 5 /log(10/2)

Which, if we used base 10 would mean that x was 0.210....
2. That's right right until the last minute. Notice that , so what is the answer?

[And as it happens, it doesn't matter which base we take the logarithm to, the RHS always evaluates as the same answer.]
3. (Original post by nuodai)
That's right right until the last minute. Notice that , so what is the answer?

[And as it happens, it doesn't matter which base we take the logarithm to, the RHS always evaluates as the same answer.]
Ofc, Log 5/ Log 5 = 1, therefore x=1. Cheers.

I was thinking because log 10 over log 2 (of which that is a simplification) = log 8 that you couldnt simply change 10/2 to 5.
4. (Original post by Jordan656)
I think I may be right, my laziness regarding learning Latex is to blame. It's not on the bottom it's Log (10/2).

Could that simplify to Log 8?
What I'm saying is that , since , and so (and it's not 0.210...).
5. (Original post by nuodai)
What I'm saying is that , since , and so (and it's not 0.210...).
yeah, got it cheers.

One thing I don't understand.

Log(10/2) is a simplification of Log 10/ Log 2

Log 10 / Log 2 = Log 8

Log (10/2) = Log 5
6. (Original post by Jordan656)
yeah, got it cheers.

One thing I don't understand.

Log(10/2) is a simplification of Log 10/ Log 2

Log 10 / Log 2 = Log 8

Log (10/2) = Log 5
Oh. In that case you sort-of got to the answer (i.e. everything up to and including log 5 / log (10/2) was correct) using a completely wrong method, so the fact you got it right up to that point was a fluke.

The laws of logs are as follows:
(i)
(ii)
(iii)
(iv)

Now, here you have . Taking logs of both sides gives:

Using (i) on the LHS gives

Using (iv) on the two terms with x as a power gives

...and then you rearrange and simplify to solve for x.

I have absolutely no idea where you're getting from... it certainly shouldn't be appearing anywhere, that's for sure! It's also certainly not true that !
7. I think you're sort of getting the second log law the wrong way around, it is not true that log(10)/log(2) = log(8) but it is true that log(10) - log(2) = log(10/2) so I can see where you got the log(8) from I think, as you thought that divide converted to a minus, rather than the other way around. Also, log(10/2) isn't "a simplification of log(10)/log(2)", it's something totally different.
8. (Original post by nuodai)
Oh. In that case you sort-of got to the answer (i.e. everything up to and including log 5 / log (10/2) was correct) using a completely wrong method, so the fact you got it right up to that point was a fluke.

The laws of logs are as follows:
(i)
(ii)
(iii)
(iv)

Now, here you have . Taking logs of both sides gives:

Using (i) on the LHS gives

Using (iv) on the two terms with x as a power gives

...and then you rearrange and simplify to solve for x.

I have absolutely no idea where you're getting from... it certainly shouldn't be appearing anywhere, that's for sure! It's also certainly not true that !
Oh, I see, I did all that, the log 8 just came from a mis understanding of rule iii above.

Self teaching Logarithms for the lose...

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