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general question on logs

Ok so i have 6 Log 3 - 99 log 3, can I write this as -93 Log 3 ?
Yes. Let x=log⁡3x=\log3. Then 6x−99x=−93x=−93log⁡3 6x-99x=-93x=-93\log3 where in the last step I replaced x back with log⁡3\log3

Another way to think of it: You can factorise log⁡3\log3 out and get (6−99)[log⁡3]=−93log⁡3 (6-99) [\log 3] = -93\log3
(edited 7 years ago)
What you could do is take the numbers in front of the log and make them the power of the numbers after the log. Seeing as both logs have the same base (10), you could use a rule to simply the expression further.
Original post by M0nkey Thunder
What you could do is take the numbers in front of the log and make them the power of the numbers after the log. Seeing as both logs have the same base (10), you could use a rule to simply the expression further.


That really is overkill....

The user should appreciate that algebra works here: If you have 6 lots of log 3 and take away 99 lots of log 3, you are left with -93 lots of log 3
Original post by Math12345
That really is overkill....

The user should appreciate that algebra works here: If you have 6 lots of log 3 and take away 99 lots of log 3, you are left with -93 lots of log 3


That's true. The questions that I do on logs usually involve having the same base numbers and wanting you apply the laws of logs, hence why I gave the answer I did :tongue:
Original post by M0nkey Thunder
What you could do is take the numbers in front of the log and make them the power of the numbers after the log. Seeing as both logs have the same base (10), you could use a rule to simply the expression further.


Original post by Math12345
That really is overkill....



His method is correct, but the only reason i'd go with @Math12345 's method is because the coefficients are slightly large to calculate as a power. Unless OP is using a calculator...
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Zacken
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Original post by NadeemKha_Arab
His method is correct


Not 'really' - it's a very circuitous proof. If one wants to get technical.
ac+bc=(a+b)c=c(a+b) for all a,b,c∈Rac + bc = (a+b)c = c(a+b)\text{ for all } a, b, c \in \mathbb{R}.

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