IB Mathemetics SL Portfolio: Logarithmic Bases

With this portfolio job, I will look into the rules of logarithms by identifying the logarithmic sequences. After determining the routine, I will create a general declaration which identifies the collection. I will after that test the validity of my basic statement by making use of other ideals. I will finally conclude the portfolio process by describing how I arrived to my personal general declaration and its limits.

Consider the next sequences. Take note of the next two terms of every sequence.

To get the initial row, another two terms of the sequence will be log64 almost 8 and log128 8. To get the second line, the next two terms of the sequence would be log243 81 and log729 seventy eight. For another row, another two the sequence can be log3125 twenty-five and log15625 25. To get the fourth line, the next two terms of the sequence would be logm5 mk and logm6 mk.

Find an expression for the nth term of each pattern. Write your expressions in the form p/q where l, q Unces. Justify your answer employing technology.

To find the nth term, I actually first resolved the series on the initial row. This really is shown under.

Log2 almost 8 = by Log4 almost 8 = back button Log8 almost eight = back button Log16 8 = x Log32 almost eight =x в‡’2x = almost 8 в‡’4x sama dengan 8 в‡’8x = eight в‡’16x = 8 в‡’32x = almost eight в‡’2x sama dengan 23 в‡’(22)x = twenty three в‡’x = 1 в‡’(24)x = twenty three в‡’(25)x = 23 в‡’x = 3 в‡’2x sama dengan 3 в‡’4x = three or more в‡’5x = 3 в‡’x = 3/2 в‡’x sama dengan 3/4 в‡’ x= 3/5

Log64 almost 8 = back button Log128 eight = back button

в‡’64x = 8 в‡’128x = almost eight

в‡’(26)x sama dengan 23 в‡’(27)x = twenty three

в‡’6x sama dengan 3 в‡’7x = several

в‡’x = 3/6 в‡’ x = 3/7

Searching at these answers intended for the initial sequence, I recently came across the fact the denominator of each and every answer displays the nth term. For instance , my answer to Log2 8 (which is definitely the first term) was 3 (or 3/1) and my answer to Log4 8 (which is the second term from the sequence) was 3/2. Searching at my answers to the initial sequence, I discovered out that the nth term could be found by the manifestation 3/n.

For the other row, We solved the answers for the sequence, which can be shown beneath.

Log3 81 sama dengan x Log9 81 sama dengan x Log27 81 = x Log81 81 sama dengan x Log243 81 = x в‡’3x = 81 в‡’9x = 81 в‡’27x = 81 в‡’81x sama dengan 81 в‡’243x = seventy eight в‡’3x = 34 в‡’(32)x = 34 в‡’(33)x sama dengan 34 в‡’x = you в‡’(35)x sama dengan 34 в‡’x = some в‡’2x = 4 в‡’3x = 5 в‡’5x = 4 в‡’x = 4/2 в‡’x = 4/3 в‡’x = 4/5

Log729 81 = back button

в‡’729x sama dengan 81

в‡’(36)x = thirty four

в‡’6x = 4

в‡’x = 4/6

By looking for these answers for the 2nd sequence, I noticed that it is just like the first pattern. Again, the denominator of each answer displays the nth term. My own answer pertaining to Log3 seventy eight (the initial term) was 4 (or 4/1) and Log9 81(the second term) was 4/2. This demonstrates that the expression pertaining to the nth term could possibly be found by 4/n

Pertaining to the third row, I fixed the answers to the series, which is demonstrated below.

Log5 twenty-five = times Log25 25 = by Log125 twenty-five = back button Log625 25 = x Log3125 25 = times в‡’5x = 25 в‡’25x = 25 в‡’125x = 25 в‡’625x = 25 в‡’3125x = 25 в‡’5x = 52 в‡’(52)x sama dengan 52 в‡’(53)x = 52 в‡’(54)x sama dengan 52 в‡’(55)x = 52 в‡’ x = two в‡’2x = 2 в‡’3x = a couple of в‡’4x sama dengan 2 в‡’5x = 2 в‡’x = 2/2 в‡’x = 2 to 3 в‡’x sama dengan 2/4 в‡’x = 2/5

Log15625 25 = Times

в‡’15625x = 25

в‡’(56)x = 52

в‡’6x sama dengan 2

в‡’x = 2/6

By looking with the answers pertaining to the third series, I found away that the denominator again reveals the nth term. The expression for the nth term would be found by 2/n

For the fourth row, My spouse and i solved the answers to the sequence, which is shown under

Logm mk = x Logm2 mk = back button Logm3 mk = by Logm4 mk = by Logm5 mk = x в‡’mnx sama dengan mk в‡’m2x = mk в‡’m3x sama dengan mk в‡’m4x = mk в‡’m5x = mk в‡’nx = t в‡’2x = k в‡’3x = k в‡’4x = k в‡’5x = t в‡’x = k/n в‡’x = k/2 в‡’x sama dengan k/3 в‡’x = k/4 в‡’x = k/5

Logm6 mk sama dengan x

в‡’m6x = mk

в‡’6x sama dengan k

в‡’x = k/6

By looking at the answers intended for the fourth collection, the design...