Logarithms: simplifying log_4^(16/x) - log_4^(16) - log4^(x)  TOPIC_SOLVED

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Logarithms: simplifying log_4^(16/x) - log_4^(16) - log4^(x)

Postby Exponential on Fri Jun 19, 2009 1:25 am

I've recently finished the Basic Log Rules and those topics that are related to exponential and logarithms. The only question I've encountered is this:

For this log:

log_4^(16/x)
- log_4^(16) - log4^ (x)

I don't get how it'll become

log_4^(16) = 2

It says refer to the basic log but I still don't get it...

Is it because 4^2 = 16???

Thanks

Second question is what I've been doing is looking at numbers. What sort of exp/log questions in words will I encounter?
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Postby stapel_eliz on Fri Jun 19, 2009 12:19 pm

Exponential wrote: log_4^(16/x)
- log_4^(16) - log4^ (x)

I will guess that the "to the power of" parts are actually the arguments of the logs, rather than powers on the bases, so the expression is as follows:

. . . . .

If so, then use a log rule to expand the first term:

. . . . .

The first and third terms cancel out, leaving you with:

. . . . .

:wink:
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Re: Logarithms: simplifying log_4^(16/x) - log_4^(16) - log4^(x)  TOPIC_SOLVED

Postby Exponential on Sat Jun 27, 2009 4:56 am

oh so this is like:

log_4^(16) - log_4^(x)

4^2 = 16

2- log_4^(x) oh that's like it. Thanks
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