Regarding logs: logx = log(17)-3 (logs are base-ten)

[bjarni86]

Hey guys!

Been working on a problem that was originally set up as logx=log(17)-3

These are all 10 base logs.

I basically started with converting the 3 to log1000 and then applying t he rule of log(A/B) = log A - log B

That leaves me with logx = log17/log1000

I am sure the answer to this is very simple... but every online calculator seems to end with x=17/1000

What happens to the logs? why would it not be x=log17/log1000 for example.

Any help very appreciated as always

 

[Deleted member 4993]

Hey guys!

Been working on a problem that was originally set up as logx=log(17)-3

These are all 10 base logs.

I basically started with converting the 3 to log1000 and then applying t he rule of log(A/B) = log A - log B

That leaves me with logx = log17/log1000..... Incorrect It should be → log(x) = log(17/1000)

I am sure the answer to this is very simple... but every online calculator seems to end with x=17/1000

What happens to the logs? why would it not be x=log17/log1000 for example.

Any help very appreciated as always

.

 

[bjarni86]

Misinterpretation of the rule... oh dear!

Thank you very much. I appreciate the assitance.

 

[bjarni86]

The question still remains how logx=log(17/1000) is converted to x=17/1000.

Is there some rule that specifies that the 2 logs of the same base cancel each other out or am I just dividing by log on both sides?

 

[bjarni86]

See pic below of how my calculations are set up so far...

The task at hand is to simplify the logarithm and give results as exact as well as approximate.

My experience tells me that an exact result would usually give you x=logx/logy and after calculating the logs, you would get the approximate decimal result.

 

[tkhunny]

Well, then you have not completed your assignment!

x =17/1000
and
x = 0.017

Both are exact. That should be okay.

 

[bjarni86]

Well, no...

What I am lacking is the understanding of how the logs cancel out (so to speak).

Can x=log(17/1000)

What determines when log expressions can be cancelled out...

 

[mmm4444bot]

The question still remains how log(x)=log(17/1000) is converted to x=17/1000.

If you use function notation, then you ought to use it consistently throughout.

Can x = log(17/1000)

Not in this exercise.

It is log(x) that equals log(17/1000)

What determines when log expressions can be cancelled out...

We not cancelling anything.

If two logarithmic expressions (same base) are equal, then their inputs (i.e., arguments) must be equal. In other words, if we know that:

log(x) = log(17/1000)

then the inputs (shown in blue) must be equal. :cool: