Good day my students, you're all welcome, today , i will teach you how to solve algorithms and it relationship with indices.Algorithms is one of those core mathematics that has to do with rules, and its operations. If you want to be the best in solving algorithms, you must be familiar with the instructions given.
There are are some rules you have to understand before thinking of solving algorithms and how to manipulate any questions based on algorithms ans indices. Something like:
n² = n x n
3² = 9 is log₃ 9 = 2
1 1
also 10⁻¹ is -------- then log₁₀ --------- = - 1
10 10
49 = 7² then log₇ 49 = 2
144 = 12² then log₁₂ 144 = 2
Note : If P² = Z hence , log p Z = 2
Now you can see, all the algorithms has their rules
to carry them during operation.Like the ones we will solve together, just pay attention you will understand them.
R² = T hence , log r T = 2 , 4² = 16 hence, log₄ 16 = 2 and so on......
Now lets simplify some algorithms
Solve the following
i. log₁₀ 100
ii. log₄ 16
iii. log₅ 125
1
iv. log --------
25
Solution
i. log₁₀ 100
Let log₁₀ 100 = y ....make sure after = is letter (from A -- Z)
Let log₁₀ 100 = x
10ˣ = 100
10ˣ = 10²
x = 2 ------ 10 and 10 is 1,so remove them to fix x and 2.
ii. log₄ 16
Let log₄ 16 = x you can change the letter x and use any letter of your choice....
4ˣ = 16
4ˣ = 4²
x = 2 ----- 4 and 4 is 1.........
iii. log₅ 125
Let log₅ 125 = x
5ˣ = 125
5ˣ = 5³
x = 3
1
iii. Let log₅ -------- = n
25
1
5ⁿ = -----------
25
1
5ⁿ = ---------
5²
5ⁿ = 5⁻²
n = - 2 remember 5⁻² is 1/5² , we just remove the 2 fives to fix n and - 2.
Relationship Between Algorithms and Indices
The operation involved in this mathematics is changing from algorithm to indices or indices to algorithms
Logₑ B = x ------- logarithms pattern .
eˣ = B ----Indices pattern
log₅ 125 = x algorithm form
5ˣ = 125 ------ indices form
Lets write the following indices in algorithm form
i. 2ˣ = 3
ii. 7² = 49
iii. 8² = 64
Solution
i. 2ˣ = 8
log₂ 8 = x
ii. 7² = 49
log ₇ 49 = 2
iii. 8² = 64
log₈ 64 = 2
Lets write the following algorithms in indices form
i.. logₑ 27 = 3
1
ii. log ₄₉ ---------
2
iii. log₁₀ x = -1
Solution
i.. logₑ 27 = 3
e³ = 27
1
ii. log ₄₉ y ---------
2
49¹⁺² = y
iii. log₁₀ x = -1
10⁻¹ = x
I believe you understand my lesson, just try and understand the common ideas behind this mathematics.
If you have any question, please put them in the comments box.
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