The
Purplemath Forums 
Logarithms:
Simplifying with Sections: Introduction to logs, Simplifying log expressions, Common and natural logs
This log is equal to some number, which I'll call y. This naming gives me the equation log_{2}(8) = y. Then the Relationship says: 2^{ y} = 8 That is, log_{2}(8), also known as y, is the power that, when put on 2, will turn 2 into 8. The power that does this is 3: 2^{3} = 8 Since 2^{ y} = 8 = 2^{3}, then it must be true that y = 3, and I get:
log_{2}(8) = 3
The Relationship says
that, since log_{5}(25)
= y, then
5^{
y}
= 25. This means
that the given log log_{5}(25)
is equal to the power y
that, when put on 5,
turns 5
into 25.
The required power is 2,
because 5^{2}
= 25. Then 5^{2}
= 5^{
y}
= 25, so: log_{5}(25) = 2
The Relationship says that this log represents the power y that, when put on 64, turns it into 4. Remembering that 4^{3} is 64, and remembering that fractional exponents correspond to roots, this means that the cube root of 64 is 4, so 64^{(1/3)} = 4. Then: log_{64}(4) = ^{1}/_{3} This last example highlights the fact that, to be able to work intelligently with logs, you need to be pretty good with your exponents. So take the time to review them, if you're feeling a little shaky.
The Relationship says that, since log_{6}(6) = y, then 6^{ y} = 6. But 6 = 6^{1}, so 6^{ y} = 6^{1}, and y = 1. That is: log_{6}(6) = 1 This is always true: log_{b}(b) = 1 for any base b, not just for b = 6.
The Relationship says that, since log_{3}(1) = y, then 3^{ y} = 1. But 1 = 3^{0}, so 3^{ y} = 3^{0}, and y = 0. That is: log_{3}(1) = 0 This is always true: log_{b}(1) = 0 for any base b, not just for b = 3.
The Relationship says that, since log_{4}(–16) = y, then 4^{ y} = –16. But wait! What power y could possibly turn a positive 4 into a negative 16? This just isn't possible, so the answer is: no solution This is always true: log_{b}(a) is undefined for any negative argument a, regardless of what the base is.
The Relationship says that, since log_{2}(0) = y, then 2^{ y} = 0. But wait! What power y could possibly turn a 2 into a zero? This just isn't possible, so the answer is: no solution This is always true: log_{b}(0) is undefined for any base b, not just for b = 2.
The Relationship says that "log_{b}(b^{3}) = y" means "b^{ y} = b^{3}". Then clearly y = 3, so: log_{b}(b^{3}) = 3 This is always true: log_{b}(b^{n}) = n for any base b. Some students like to think of the above simplification as meaning that the b and the logbaseb "cancel out". This is not technically correct, but it can be a useful way of thinking of things. Just don't say it out loud in front of your instructor.
Remember that a logarithm is just a power; it's a lumpy and long way of writing the power, but it's just a power, nonetheless. The expression "log_{2}(9)" means "the power which, when put on 2, turns 2 into 9." And they've put that power onto 2, which means that the 2 has been turned into 9! Looking at it another way, "2^{log2(9)} = y" means "log_{2}(y) = log_{2}(9)" (which is the equivalent logarithmic statement), so y = 9. But y = 2^{log2(9)}, so 2^{log2(9)} = 9. While the second way is technically correct, I find the first way to be more intuitive and understandable. Either way, though, I get an answer of: 2^{log2(9)} = 9 This last example probably looks very complicated, and, in the technical explanation, it is. Look instead at the intuitive explanation (in the first paragraph). Some students even view the above problem as the 2 and the logbase2 as "cancelling out", which is not technically correct, but can be a useful way of remembering how this type of problem works. To synopsize, these are the things you should know from this lesson so far:
<< Previous Top  1  2  3  Return to Index Next >>



Copyright © 20022012 Elizabeth Stapel  About  Terms of Use 




