Logarithm

Definition Logarithm :
If m= a ^x, where a>0 ,a not equal to 1 , m>0 then x is called the logarithm of m to the base a and is written as x = log _a m, where m, a, x are real numbers.

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Laws of logarithms:
1. Law of Product: log (A.B) =log A +log B
2. Law of Quotient : log(A/B) =log A - log B
3. Law of Exponent : log m ⁿ = n log m

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Properties:
1)We have m = a ^x if and only if x = log _a m.
2)Logarithms of negative numbers and zero are not defined.
3) a⁰ =1 Hence, 0 = log _a 1
i.e. logarithm of 1 to any base is always zero.
4) a¹ =a Hence , 1 = log _a a
i.e. logarithm of any number to the same base is 1.
5) x = log _a a^x
6) If log _a m = log _a n then m = n
7)If a>1 and m>n then log _a m > log _a n and conversely.

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Write logarithm form:
1) 5 ³ = 125
Solution:
3 = log ₅ 125   (By definition of logarithm)

2) 8 ³ =512
Solution:
3 = log ₈ 512   (By definition of logarithm)

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Express in exponential form:
1) log ₂ 128 = 7
Solution: 128 = 2 ⁷  (By definition of logarithm)

2) log ₉ 6561 =4
Solution: 6561 = 2 ⁹  (By definition of logarithm)

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Simplify : 
1) log 5 + 2 log 4
Solution:
=log 5 + log 4² (By Law of Exponent)
=log 5 + log 16
=log (5 . 16) (By law of Product)
=log 80

2) 2 log 7 - log 14
Solution:
=log 7 ² - log 14 (By Law of Exponent)
=log 49 - log 14
=log (49/14) (By law of Quotient)
=log (7/2)

3)log 3 + log 2 - 2 log 5
Solution:
=log 3 + log 2 - log 5² (By Law of Exponent)
=log 3 + log 2 - log 25
=log (3 . 2) - log 25 (By Law of Product)
=log 6 - log 25
=log (6/25) (By Law of Quotient)

4) 2 log 3 - 1/2 log 16 + log 12
Solution:
=log 3 ² - log 16 ^1/2 + log 12 (By Law of Exponent)
=log 9 - log4 + log 12
=log (9/4) + log 12 (By Law of Quotient)
=log (9/4 .12) (By Law of Product)
=log (9. 3)
=log 27

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Prove that:
1) log 540 = 2 log 2 + 3 log 3 + log 5
Solution:
L.H.S = log (540)
=log (4 X 27 X 5)
=log ( 2² X 3³ X 5)
=log 2² + log 3³ + log 5 (By Law of Product)
=2 log 2 + 3 log 3 +log 5 (By Law of Exponent)
=R.H.S.
Hence, log 540 = 2 log 2 + 3 log 3 + log 5

OR

R.H.S =2 log 2 + 3 log 3 + log5
=log 2 ² + log 3³ + log 5 (By Law of Exponent)
=log 4 + log 27 + log 5
=log (4 X 27 X 5) (By Law of Product)
=log 540
=L.H.S.
Hence, log 540 = 2 log 2 + 3 log 3 + log 5

2) log 360 = 3 log 2 + 2 log 3 + log 5
R.H.S =3 log 2 + 2 log 3 + log5
=log 2 ³ + log 3² + log 5 (By Law of Exponent)
=log 8 + log 9 + log 5
=log (8 X 9 X 5) (By Law of Product)
=log 360
=L.H.S.
Hence, log 360 = 3 log 2 + 2 log 3 + log 5

OR

L.H.S = log (360)
=log (8 X 9 X 5)
=log ( 2³ X 3² X 5)
=log 2³ + log 3² + log 5 (By Law of Product)
=3 log 2 + 2 log 3 + log 5 (By Law of Exponent)
=R.H.S.
Hence, log 360 = 3 log 2 + 2 log 3 + log 5

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Evaluate:
1) log 81/log 9
=log 9² / log 9
=2 log 9 / log 9   (By Law of Exponent)
=2

2)log ₂ 5 / log₂ 11 - log₄ 5 / log₄ 11
Solution :
We know ,Change of Base formula is log _a m = log m /log a

=log 5/log 2/log 11/log 2 - log 5/log 4/ log 11 /log 4
=log 5 . log 2 / log 11 . log 2 - log 5 .log 4 / log 11 .log 4
=log 5 / log 11 - log5 / log 11
=0

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Embedded systems

Difference between Static RAM Dynamic RAM.

Static RAM	Dynamic RAM
1. SRAM	1. DRAM
2. It retains it levels so no require refreshing	2. It requires some refreshing.
3. Power consumption is more	3. Power consumption is less.
4. Speed is fast.	4. Speed is less.
5. Package density is low.	5. Package density is high.
6. Size can't be increase.	6. Size can increase.
7. More expensive.	7. Less expensive.
8. Made up of 6 CMOS transistors(MOSFET).	8. Made up of a MOSFET and a capacitor.

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Difference between RISC and CISC processors/controllers.

RISC	CISC
1. It stands for Reduced Instruction Set Computing.	1. It stands for Complex Instruction Set Computing.
2. Lesser number of instructions.	2. Greater number of Instructions
3. Instruction pipe-lining and increased execution speed.	3. Generally no instruction pipe-lining feature.
4. Orthogonal instruction set.	4. Non-Orthogonal instruction set.
5. Operations are performed on register only, the only memory operations are load and store.	5. Operations are performed on register or memory depending on the instruction.
6. A large number of registers are available	6. Limited number of general purpose register.
7. Programmer needs to write more code to execute a task since the instruction are simpler ones.	7. Instructions are like macros in C language. A programmer can achieve the desired functionality with a single instruction.
8. Single, fixed length instructions.	8. Variable length instructions
9. Less silicon usage and pin count.	9. More silicon usage since more additional decoder logic as required.
10. With Harvard Architecture.	10. Can be Harvard or Von- Neumann Architecture.

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Difference between microprocessor and micro-controller.

Microprocessor	Micro-controller
1.A silicon chip representing a central processing unit (CPU), which is capable of performing arithmetic as well as logical operations according to a pre defined set of instructions.	1.A micro-controller is a highly integrated chip that contains a CPU, scratchpad RAM, special and general purpose register arrays, on chip ROM / FLASH memory for program storage, timer and interrupt control units and dedicated  I/O ports.
2.It is a dependent unit .It requires the combination of other chips like timers, program and data memory chips, interrupt controllers, etc. for functioning.	2.It is a self-contained unit and it doesn't require external interrupt controller , timer , UART , etc. for its functioning.
3.Most of the time general purpose in design and operation .	3.Mostly application-oriented or domain - specific.
4. Doesn't, contain a built in I/O port.The I/O port functionality needs to be implemented with the help of external programmable peripheral interface chips like 8255.	4.Most of the processor contain multiple built-in I/O ports which can be operated as a single 8 or 16 or 32 bit port or as individual port pins.
5.Targeted for high end market where performance is important.	5.Targeted for embedded market where performance is not critical.
6.Limited power saving options compared to micro-controllers.	6.Includes lot of power saving features.

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Difference between Harvard Architecture and Von-Neumann Architecture.

Harvard Architecture	Von-Neumann Architecture.
1.Separate buses for instruction and data fetching.	1.Single shared bus for instruction and data fetching.
2.Easier to pipeline , so high performance can be achieved .	2. Low performance compared to Harvard architecture
3.Comparatively high cost.	3. Cheaper
4.No memory alignment problems.	4. Allow self modifying codes.
5.Since data memory and program memory are stores physically in different locations, no chances for accidental corruption of program memory.	5. Since data memory and program memory are stores physically in the same chip , chances for accidental corruption of program memory.

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