LOGARITHMS

Log Rules:


1) log_b(mn) = log_b(m) + log_b(n)


2) log_b(^m/_n) = log_b(m) – log_b(n)


3) log_b(mⁿ) = n · log_b(m)

In less formal terms, the log rules might be expressed as:

1) Multiplication inside the log can be turned into addition outside the log, and vice versa.


2) Division inside the log can be turned into subtraction outside the log, and vice versa.


3) An exponent on everything inside a log can be moved out front as a multiplier, and vice versa.

Warning: Just as when you're dealing with exponents, the above rules work only if the bases are the same. For instance, the expression "log_d(m) + log_b(n)" cannot be simplified, because the bases (the "d" and the "b") are not the same, just as x² × y³ cannot be simplified (because the bases x and y are not the same).

Expanding logarithms

Log rules can be used to simplify expressions, to "expand" expressions, or to solve for values.

• Expand log₃(2x).

When they say to "expand", they mean that they've given you one log expression with lots of stuff inside it, and they want you to use the log rules to take the log apart into lots of separate logs, each with only one thing inside. That is, they've given you one log with a complicated argument, and they want you to convert this to many logs, each with a simple argument.

I have a "2x" inside the log. Since "2x" is multiplication, I can take this expression apart and turn it into an addition outside the log:


log₃(2x) = log₃(2) + log₃(x)


The answer they are looking for is:


log₃(2) + log₃(x)

Do not try to evaluate "log₃(2)" in your calculator. While you would be correct in saying that "log₃(2)" is just a number, they're actually looking here for the "exact" form of the log, as shown above, and not a decimal approximation from your calculator.

• Expand log₄( ¹⁶/_x ).

I have division inside the log, which can be split apart as subtraction outside the log, so:


log₄( ¹⁶/_x ) = log₄(16) – log₄(x)


The first term on the right-hand side of the above equation can be simplified to an exact value, by applying the basic definition of what a logarithm is:


log₄(16) = 2


Then the original expression expands fully as:


log₄( ¹⁶/_x ) = 2 – log₄(x)

Always remember to take the time to check to see if any of the terms in your expansion (such as thelog₄(16) above) can be simplified.

• Expand log₅(x³).   Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved'

The exponent inside the log can be taken out front as a multiplier:



log₅(x³) = 3 · log₅(x) = 3log₅(x)

Expand the following:   

        

The 5 is divided into the 8x⁴, so split the numerator and denominator by using subtraction:

Don't take the exponent out front yet; it is only on the x, not the 8, and you can only take the exponent out front if it is "on" everything inside the log. The 8 is multiplied onto the x⁴, so split the factors by using addition:



log₂(8x⁴) – log₂(5) = log₂(8) + log₂(x⁴) – log₂(5)


The x has an exponent (which is now "on" everything inside its log), so move the exponent out front as a multiplier:



log₂(8) + log₂(x⁴) – log₂(5)


    = log₂(8) + 4log₂(x) – log₂(5)


Since 8 is a power of 2, I can simplify the first log to an exact value:



log₂(8) + 4log₂(x) – log₂(5)


   = 3 + 4log₂(x) – log₂(5)


Each log contains only one thing, so this is fully simplified. The answer is:



3 + 4log₂(x) – log₂(5)


• Expand the following:
           

Use the log rules, and don't try to do too much in one step:




Then the final answer is:   C
opyright © Elizabeth Stapel 2002-2011 All Rights Reserved
log₃(4) + 2log₃(x – 5) – 4log₃(x) – 3log₃(x – 1)




• Simplify log₂(x) + log₂(y).
Since these logs have the same base, the addition outside can be turned into multiplication inside:



log₂(x) + log₂(y) = log₂(xy)


The answer is log₂(xy).


• Simplify log₃(4) – log₃(5).
Since these logs have the same base, the subtraction outside can be turned into division inside:



log₃(4) – log₃(5) = log₃(⁴/₅)


The answer is log₃(⁴/₅).


• Simplify 2log₃(x). Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved
The multiplier out front can be taken inside as an exponent:

2log₃(x) = log₃(x²)


• Simplify 3log₂(x) – 4log₂(x + 3) + log₂(y).
I will get rid of the multipliers by moving them inside as powers:


3log₂(x) – 4log₂(x + 3) + log₂(y) 

     = log₂(x³) – log₂((x + 3)⁴) + log₂(y)


Then I'll put the added terms together, and convert the addition to multiplication:


log₂(x³) – log₂((x + 3)⁴) + log₂(y) 

     = log₂(x³) + log₂(y) – log₂((x + 3)⁴) 

     = log₂(x³y) – log₂((x + 3)⁴)


Then I'll account for the subtracted term by combining it inside with division:




• Evaluate log₃(6).
The argument is 6 and the base is 3. I'll plug them into change-of-base, using the natural log as my new log:

Then the answer, rounded to three decimal places, is:



log₃(6) = 1.631


You would get the same answer if you used the common log, though the numerator and denominator of the intermediate fraction would be different from what I did above:




As you can see, it doesn't matter which standard-base log you use, as long as you use the same base for the numerator and denominator.   Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved
While I showed the numerator and denominator values in the above calculations, it is actually best to do the calculations entirely within your calculator. You don't need to bother writing out the intermediate step. To minimize on round-off errors, try to do all the steps for the division and evaluation in your calculator, all in one go. In the above computation, rather than writing down the first eight or so decimal places in the values of ln(6) and ln(3) and then dividing, you would just do "ln(6) ÷ ln(3)" in your calculator.
You may also get some simple (but useless) exercises on this topic, such as:


• Convert log₃(6) to base 5.
I can't think of any particular reason why a base-5 log might be useful, so I think the only point of these problems is to give you practice using change-of-base:


• Convert ln(4) to an expression written in terms of the common log.
Again, why would you do this (in "real life"), since you can already evaluate the natural log in your calculator? You wouldn't; this exercise is just for practice:


Since getting an actual decimal value is not the point in exercises of this sort (the converting using change-of-base is the point), just leave the answer as a logarithmic fraction.


On the other hand, using change-of-base is handy for finding plot-points when graphing non-standard logs, especially when you are supposed to be using a graphing calculator.


• Use your graphing utility to graph y = log₂(x).
If you were working by hand, you would use the definition of logs to note that:

• since 2^–1 = ¹/₂, then log₂(1/2) = –1
• since 2⁰ = 1, then log₂(1) = 0
• since 2¹ = 2, then log₂(2) = 1
• since 2² = 4, then log₂(4) = 2
• since 2³ = 8, then log₂(8) = 3
• since 2⁴ = 16, then log₂(16) = 4