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Logarithms: Simplifying with
"The Relationship"
(page 2 of 3)

Sections: Introduction to logs, Simplifying log expressions, Common and natural logs

• Simplify log2(8).

This log is equal to some number, which I'll call y. This naming gives me the equation log2(8) = y. Then the Relationship says:

2 y = 8

That is, log2(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:

23 = 8

Since 2 y = 8 = 23, then it must be true that y = 3, and I get:

log2(8) = 3

• Simplify log5(25).

The Relationship says that, since log5(25) = y, then 5 y = 25. This means that the given log log5(25) is equal to the power y that, when put on 5, turns 5 into 25. The required power is 2, because 52 = 25. Then 52 = 5 y = 25, so:

log5(25) = 2

The Relationship says that this log represents the power y that, when put on 64, turns it into 4. Remembering that 43 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:

log64(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.

• Simplify log6(6).

The Relationship says that, since log6(6) = y, then 6 y = 6. But 6 = 61, so 6 y = 61, and y = 1. That is:

log6(6) = 1

This is always true: logb(b) = 1 for any base b, not just for b = 6.

• Simplify log3(1).

The Relationship says that, since log3(1) = y, then 3 y = 1. But 1 = 30, so 3 y = 30, and y = 0. That is:

log3(1) = 0

This is always true: logb(1) = 0 for any base b, not just for b = 3.

• Simplify log4(–16).

The Relationship says that, since log4(–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: logb(a) is undefined for any negative argument a, regardless of what the base is.

• Simplify log2(0).

The Relationship says that, since log2(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: logb(0) is undefined for any base b, not just for b = 2.

• Simplify logb(b3).

The Relationship says that "logb(b3) = y" means "b y = b3". Then clearly y = 3, so:

logb(b3) = 3

This is always true: logb(bn) = n for any base b.

Some students like to think of the above simplification as meaning that the b and the log-base-b "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.

• Simplify 2log2(9).

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 "log2(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, "2log2(9) = y" means "log2(y) = log2(9)" (which is the equivalent logarithmic statement), so y = 9. But y = 2log2(9), so 2log2(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:

2log2(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 log-base-2 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:

• The Relationship: "logb(x) = y" means the same thing as "b y = x".
• Logarithms are really exponents (powers); they're just written differently.
• logb(b) = 1, for any base b, because b1 = b.
• logb(1) = 0, for any base b, because b0 = 1.
• logb(a) is undefined if a is negative.
• logb(0) is undefined for any base b.
• logb(bn) = n, for any base b.

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 Cite this article as: Stapel, Elizabeth. "Logarithms: Simplifying with 'The Relationship'." Purplemath. Available from     http://www.purplemath.com/modules/logs2.htm. Accessed [Date] [Month] 2014

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