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Mathematics 16 Online
OpenStudy (anonymous):

If log(17) + log(y) = log(z), then 17 + y = z. true or false? i thought z would be 17*y

OpenStudy (anonymous):

since its addition it goes with multiplication

OpenStudy (anonymous):

Yes you think right..

OpenStudy (anonymous):

so the answer would be false

OpenStudy (anonymous):

\[\log(a) + \log(b) = \log(ab)\]

OpenStudy (anonymous):

what about: If ln(17x) = ln (y), then 17x = y.

OpenStudy (anonymous):

Yes it will be false.. 17y = z..

OpenStudy (anonymous):

Yes you are right again..

OpenStudy (anonymous):

so this is false too?

OpenStudy (anonymous):

How?? \(log(a) = log(b) \implies a = b\)

OpenStudy (anonymous):

\[\log(a) = \log(b) \rightarrow a = b\]

OpenStudy (anonymous):

ah i gotcha!

OpenStudy (anonymous):

So is it false or true ??

OpenStudy (anonymous):

True

OpenStudy (anonymous):

Yes you are also true..

OpenStudy (anonymous):

Which of the following is equivalent to ln(3e^6)? 18 ln(3) + 6 36 6ln(3) 3e^6

OpenStudy (anonymous):

Firstly use : \[\log(ab) = \log(a) + \log(b)\]

OpenStudy (anonymous):

a = 3 and b = \(e^6\) here..

OpenStudy (anonymous):

dude your so helpful, i guess i just get confused when they throw an e in there but if u just look at the basic properties u can tell what to do with most of these

OpenStudy (anonymous):

(log(x))^7 = 7log(x) so this would be true, am i correct?

OpenStudy (anonymous):

Yes it is correct but : You should remember this : \[\ln(e) = 1\]

OpenStudy (anonymous):

\[\ln(e^6) = 6 \cdot \ln(e) \rightarrow 6 \cdot 1 \rightarrow \color{green}{6}\]

OpenStudy (anonymous):

Which of the following is equivalent to ln(13 + e8)? ln(13) + 8 13 + e8 104 8ln(13 + e) None of the above

OpenStudy (anonymous):

What do you think tell me first..

OpenStudy (anonymous):

i thought it was ln(13) + 8, but its not

OpenStudy (anonymous):

That is e^8 ???

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

There is no particular formula for \(log(a + b)\)..

OpenStudy (anonymous):

Option e> looks fine to me for this question..

OpenStudy (anonymous):

none of the above?

OpenStudy (anonymous):

If this is option e> then yes..

OpenStudy (anonymous):

e> ?? im confused

OpenStudy (anonymous):

?????

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