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plz tell me about the rules of logarithm to solve the questions in class 12th?

plz tell me about the rules of logarithm to solve the questions in class 12th?

Top 2 Answers
mizo

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look i finish that stage but may i help you :

Rules of Exponents.

1

ak = a−k ak an = ak+n ak

an = ak−n a

b k

= ak

bk (ak)n = akn kpa = a1/k

Rewrite each of the following expressions in the form a b c .

1

a7 b2

a b c

2 at b5

cr a2 c3

b2 3

a2 b−2 pc

a3/2 b−3 c5 4 a3pb

c7 !5

Exponential and Logarithmic Functions.

A logarithm is the inverse of an exponential. That is, loga ax = x for any positive a 6= 1, and aloga x = x.

We usually use a base of e, which is natural constant (that is, a number with a letter name, just like ).

The number e is approximately 2.7182818284590452354. The logarithm we usually use is log base e, written

loge(x) or (more often) ln(x), and called the natural logarithm of x.

Rules of Logarithms.

• Definition: c = logb(a) () a = bc

• The Big One: ln(xy) = y · ln(x) or loga(xy) = y · loga(x)

• Others: loga(r · s) = loga(r) + loga(s) loga(r/s) = loga(r) − loga(s)

loga(b) =

logx(b)

logx(a) , for any x

Solve for t (algebraically, not numerically) in the following equations.

5 200 = 5 t3 6 800 = 4 · 7t 7 400 = 200 + 3 · 2t 8 432 = 100e0.6t

9 log2(t) = 6 10 ln(t2) = 30

Functions of Exponential Type.

A function is said to be of exponential type if it can be written in the form

y = a · bt where a and b are constants.

If we are given two data points, (two pairs of t and y values), we can determine the constants a and b by

solving a system of two equations.

Example: Given that 200 = a · b2 and 450 = a · b7, we divide the second equation by the first to

get:

450

200

= b7

b2 and so 9/4 = b5, giving b = 5p9/4.

Substituting that into the first equation gives a =

200

(9/4)2/5 = 200 · (9/4)−2/5.

Find a and b given that:

11 30 = a · b5 and 80 = a · b9 12 1.5 = a · b24 and 2.3 = a · b36

(1) a6 b1 c−1 (2) at+2 b3 c3−r (3) a1/2 b1 c−9/2 (4) a15 b5/2 c−35 (5) t = 3 p40 (6) t = log7 (200) (7) t = log2 (200/3) (8)

t = (ln 4.32) /0.6 (9) t = 26 (10) t = e15 (11) b = (8/3)1/4 , a = 30 · (8/3)−5/4 (12) b = (2.3/1.5)1/12 , a = (1.53 )/(2.32 ) .

if you didn’t under stand A word log into that:

(http://www.math.toronto.edu/mathnet/SOAR2003/Winter/PDF/logs.pdf)

0

Anonymous
Not enough room here for a complete explanation.

Logarithms are exponents, powers of a base number. For example 100 can be written as 10^2 (10 squared or 10 x 10). So, the Logarithm of 100 using 10 as a base is 2. This is written as:

Log 100 = 2

Because 10 squared = 100.

(The base number is assumed to be 10 if it isn’t written)

So, log 1000 = 3

and log 7 = .845 because 10^.845 = 7

Using logarithms simplifies some mathematic operations.

If you add logs you are multiplying the numbers they represent. When you subtract them, you are dividing the numbers they represent.

So: 100 * 7 = 700 which can be expressed as:

10^2 + 10^.845 or 10^(2 + .845) = 10^2.845

so log 100 + log 7 = 2.845

I’m sure that your text book will have a very complete explanation. I suggest you take a look at it from the beginning and work some of the problems.

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