__Definition and Basics of Logarithms--Part II__Saying that *log _{a}M =x *means exactly the same thing as saying

*a*

^{x}= M .In other words:

*log*is the number to which you raise_{a}M*a*in order to get*M.*

Keep this in mind in thinking about logarithms. It makes lots of things obvious.

For example: What is *log _{2} 8? Ask yourself "To what power should I raise 2 in order to get 8?" *Since 8 is 2

^{3 }the answer is "3." So

*log*

_{2}8=3.Here's another way that remembering the rule:

*"log*is the number to which you raise_{a}M*a*in order to get*M."*

can make some things almost obvious. For example, what is *2 ^{log}*

*2*Note that

^{ 5}?*is the power to which 2 is being raised.*

^{log}_{2}^{ 5 } But * ^{log}_{2}^{ 5 }* is the number to which you raise 2 in order to get 5!

*So if you raise 2 to that number you get 5!!*In other words

*2 ^{log}2^{ 5 }= 5.*

Let's use

*"log*is the number to which you raise_{a}M*a*in order to get*M,"*

to understand logarithms of product. For example: What is* log _{2}(8*32)*?

Notice that *8=2 ^{3 }*and 32

*=2*so 8*32

^{5 }*=2*=

^{3}2^{5 }*2*.

^{3+5 }=2^{8 }But this means that

*log _{2}(8*32)=log_{2}(2^{8}) = 8 = 3+5=*

*log _{2}(2^{3})+log_{2}(2^{5})=log_{2}(8)+log_{2}(32)*

*In other words, the l og of the product 8*32 equals the sum of the logs of 8 and 32.*

Of course there is nothing special about the base 2. The same idea holds for other logarithms.

Apply this idea to the following examples: