Log5 25, Log25 25, Log125 25, Log625 25, Log312525, Log1562525, …
Considering all of the above sequences, a further more generalized way for expressing the above sequences, can be written as:
Logmmk, Logm2mk, Logm3mk, Logm4mk, ...
The nth term for this sequence is Logmnmk. Analyzing this sequence the next two terms are going to be Logm5mk and Logm6mk, which therefore, makes this sequence:
Logmmk, Logm2mk, Logm3mk, Logm4mk, Logm5mk, Logm6mk, …
Logs can also be written in other ways as well. There are also many rules that logarithms can follow. In the form of . For example, you can re-arrange the log function to become a fraction type of function such as, Log28, this can be written as . Therefore, when completing each expression given above for the sequences in the form of , the nth term of the first sequence, Log2n23, is written as:
Log2n23 =
Furthermore, there is a rule of the logarithm stating that the power is multiplied by the whole thing and will then enable the logs to cancel out. This rule is shown below:
Therefore for expression , this simplifies to become:
, This can also be simplified as the (Log 2) in both the denominator and the numerator can cancel each other out. This expression is then simplified into the expression of .
For the second sequence shown above, the nth term for that expression is Log3n34.
Log3n34 = this simplifies to become:
, this can further be simplified as the (Log 3) in the fraction, can be canceled out to further enhance the expression. This means that the expression will be shown as .
As for the third sequence, the nth term expression is known as Log5n52. This means that when writing it in the form of the expression will be shown as:
Log5n52 = , this simplifies to become:
Again, knowing that the (Log 5) will be cancelled out, the expression then becomes .
Lastly, for the generalized expression sequence, Logmmk, Logm2mk, Logm3mk, Logm4mk, Logm5mk, Logm6mk, … the nth expression that for this sequence is Logmnmk. When writing this in the form, this expression looks like:
Logmnmk = , this further simplifies to:
, looking at this fraction, this can obviously be simplified again to create a further improved expression because the (Log m) cancel out within the fraction. This expression is, .
Knowing all of the fully enhanced expressions for each sequence the expressions consist of:
These expressions basically say that each of the expressions are equal to the log form expression of the sequence. Proving this:
= Log2n23.
= Log3n34.
= Log5n52.
= Logmnmk.
To further justify this, graphing both the enhanced expression, for example , and the Log expression, Log2n23, for each sequence, the graphs should look the same. If the graph’s are identical, than this would verify that this statement is valid and correct, and it further proves that the enhanced expression is correct. All of the above expressions can be graphed except for the generalized expression, because of the fact that there are two unknowns within the expression. Also you can’t graph this expression because there aren’t enough integers to plot a graph of those expressions.
The graph below represents a graph of . (y = )
The graph below represents a graph of Log2n23. (y = Log2n23)
Analyzing the graphs shown above, you can see that both of the graphs are exactly identical. This fully justifies that both of these equations are exactly the same, and it proves that the expressions for the sequence is equal as well. They are both equal.
The graph below represents a graph of . (y = )
The graph below represents a graph of Log3n34, where y = of Log3n34.
Analyzing each graph, this is also a justification to the main expression because this follows the concept that the expressions defining the sequences are equal.
The graph below represents a graph of . (y = )
The graph below represents a graph of Log5n52, where y = of Log5n52.
Looking at these two graphs, again this is obvious that the graphs are identical. Therefore, this proves that the expressions/formulas for the sequence are equal and is valid. Therefore, this means that this is justified and proved.
This method of writing the expression for the log in this form, , can then be a way to calculate the actual answer of the Log. The following Logs are answered in this form below:
Log4 64, Log8 64, Log32 64, …
Log4 64 =
or Log4 64 = Log2226
=
= = = 3
Log8 64 =
or Log8 64 = Log2326
=
= = = 2
Log32 64 =
or Log8 64 = Log2526
=
= = = 1.2
One of the rules of logarithms is that Logb A = . This means that to find the answer to the next term in the sequence, which is Log3264. Call Log464 S1 and Log864 S2. To find the general formula you must do the following:
Since Log3264 =
This becomes, because 4 × 8 = 32 and this is the same thing.
Then, , to figure out the general formula you can substitute letters to show the main point. This means subbing in S1 and S2 because these are the first 2 terms.
This looks like, this further can simplified when getting a common denominator, = . This can be simplified as the lowest denominator moves to the numerator.
This means the general formula looks like: .
Also above in the sequences you can notice that that the power of the base, when it is simplified completely, is always the denominator of the fractional answer, and the power of the exponent when it is simplified completely is always the numerator. Another way to do answer the log above is: = this is a valid rule and statement for solving these logs. You can notice that the value for the numerator is always a constant because the exponent is always 6, so therefore 3 × 2 = 6. These number values come from the sequence, the answer to Log464 = 3 and Log864 = 2; this is where the values come from.
The next sequence is
Log7 49, Log49 49, Log343 49, …
Log7 49 =
or Log7 49 = Log7172
=
= = = 2
Log49 49 =
or Log49 49 = Log7272
=
= = = 1
Log343 49 =
or Log343 49 = Log7372
=
= =
The two values of the logs before Log34349 are 2 and 1. This means that Log34349 = , using the other method this means that = which equals .
The third sequence is:
Log1/5 125, Log1/125 125, Log1/625 125, …
Log1/5 125 = =
This equals, = = -3
Log1/125 125 = =
= This equals, = = -1
Log1/625 125 = =
= This equals, =
Using the other form, to find Log1/625 125, you must:
== =
The fourth sequence is:
Log8 512, Log2 512, Log16 512, …
Log8 512 = =
This equals, = = 3
Log2 512 = =
= This equals, = = 9
Log16 512 = =
= This equals, =
Using the other form, to find Log16 512, you must:
= =
Letting Loga X = c and Log b X = d, we can further create a general statement similar to the 2nd way of calculating the logs. (Shown above) ().
Finding the general statement that expresses Logab X, in terms of c and d is shown in the following steps:
-
Firstly we know that because of the rules of logarithms, =
-
The second step is to simplify this formula into its simplest form and you should have the complete general statement that expresses Logab X, in terms of c and d.
= = = = = .
Arriving at this general statement relates heavily to the fact about the rules of logarithms. These rules help find the final general statement that expresses the each term and answer for each term.
The Scope and Limitations:
There are a couple of limitations to this general statement. First is that for the base of all logarithms, you cannot have a negative integer for these values. This means that for a and b the numbers cannot be negative. Furthermore, this shows means that x, a and b must be greater than zero. Another way of expressing this is that the domain of this general statement must be all positive real numbers.
