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Bases to the Power of Logs 
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#1
May508, 11:10 AM

P: n/a

In class, I had a question about how to solve the following problem:
6^log6_3 (EDIT: if this is ambiguous without the math formatting, (I don't know how to format it) it reads 6 to the power of log base 6 that yields 3) No one knew how to solve it (including the teacher), and he could only tell us the answer is 3 (since he has a teachers edition with all the answers). So can anyone tell me how to do this? It's going to eat at me all day if I don't figure it out. Thanks! 


#2
May508, 11:15 AM

P: 443

6^x and Log base 6 of x are inverse functions, therefore their composite just equals x
If that doesn't answer your question try to follow these steps (remember that Logx_y^z = z*Logx_y 6^(Log6_3) = x Take the log base six of both sides Log6_(6^(Log6_3)) = Log6_x Now using that power rule I mentioned we get (Log6_3)*(Log6_6) = Log6_x Since Log6_6 = 1 we have Log6_3 = Log6_x Now do you see why x must = 3? 


#3
May508, 03:32 PM

P: n/a

Diffy, that was exactly what I needed. Quite frankly, I had no idea one could 'take the log' of an equation (but after examining it, it seemed like a valid manipulation). Thanks a bunch!



#4
May508, 03:41 PM

P: 443

Bases to the Power of Logs
Just make sure when you take the log of both sides, everything on each side gets put into the log.
EX. 6^x + 4x  sin(x) = log6_x If you were to take the log (base six) of both sides it would result in the following: log6_(6^x + 4x  sin(x)) = log6_(log6_x) similarly you can "raise" each side of an equation. Lets say you had log6_(3x2) = 2 then you can raise each side of the equation with base six like this: 6^(log6_(3x2)) = 6^(2) Again since 6^x and log6_x are inverse functions, they (in a sense) cancel and we are left with: 3x2 = 36 which makes solving for x nice and easy. 


#5
May508, 04:42 PM

P: 355

For difficult problems, it is not expected that the teacher will always know how to solve it immediately (or be able to solve it at all without a lot of work in some cases), and it's expected that even on easy problems, the teacher will sometimes have difficulty solving them if the method of solving it is not obvious (or if the teacher is tired, sick, distracted, etc); teachers are human too. However, that 6^(log_6(x)) = x (for x > 0) is the very definition of log_6. It's not odd that the teacher couldn't immediately give you the answer; he's human, everyone makes mistakes. However, it is odd that the teacher couldn't tell you why the answer was 3 after seeing the answer. 


#6
May508, 04:44 PM

P: n/a

Okay, so let me get this straight. You're saying one can have an equation logb_X = y, and, with any base B, have B^(logb_X) = B^y?
Color me ignorant, but this seems to just complicate things, to me. In your example, log6_3x2 = 2, wouldn't it just be simpler to let 3x2 = A, then say that log6_(A) = 2, and 6 to the power of 2 equals what? Since it's 36, 3x2 = 36, then solve. Don't get me wrong, I think it's pretty cool one can raise an equation to a power, but isn't this way more efficient? 


#7
May508, 04:54 PM

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#8
May508, 04:59 PM

P: 355




#9
May508, 05:10 PM

P: n/a

One last question about logarithms:
How does one find the logarithm of logb_X when X is not b^n? For example, log2_8 = 3 obviously, because 2^3 = 8. But what about log2_7 = ? How would you solve that? 7 is not a 'power' of the base 2 like 8 is. I suppose I could guess it's .98 or whatever, and I can put that into my calculator, but other than that, I'm clueless as to how one would find the log of that. 


#10
May508, 06:17 PM

P: 355

For instance, for x between 1 and 1 (exclusive, i.e., it doesn't work if x = 1 or if x = 1), a formula that you can use for the natural log (log base e) is Log(1+x) = x  x^2/x + x^3/x  x^4/4 + x^5/5  ... (and so on) since x is between 1 and 1, the terms you are adding and subtracting keep getting smaller, so you can stop after some point and be confident that your answer is pretty close (though this depends on how close x is to 0, for x closer to 1 or 1, you'd want to keep going until you get to around x^30/30) A more complicated formula that you can use and not have to take it to as many terms (though i believe that it's only valid for 0 < x < 2) is [tex]\ln (z) = 2 \sum_{n=0}^\infty \frac{1}{2n+1} {\left ( \frac{z1}{z+1} \right ) }^{2n+1}[/tex] 


#11
May508, 06:22 PM

P: 352

When I was learning about logarithms and I didn't have a good intuitive understanding of what they meant I would read e.g. [tex]\log_6 3[/tex] as "the power you have to raise six to to get three." So [tex]6^{\log_6 3}[/tex] is simply "six to the power you have to raise six to to get three." It's then obvious that this is equal to three.



#12
May608, 06:19 AM

P: 443

Think about it however you want, in my example I was offering a simple equation, so perhaps it is more efficient your way. You are solving it, basically in your head, I was showing the actual mathematical steps to go through in order to solve. If you continue with your maths there will be more difficult problems that you perhaps will not be able to solve in your head and then be forced to go through the steps. 


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