Log class 9

 Express 53 = 125 in logarithm form.

Solution:

53 = 125

As we know,

ab = c ⇒ logac=b

Therefore;

Log5125 = 3

2. Express log101 = 0 in exponential form.

Solution:

Given, log101 = 0

By the rule, we know;

logac=b ⇒ ab = c

Hence,

100 = 1

3. Find the log of 32 to the base 4.

Solution: log432 = x

4x = 32

(22)x = 2x2x2x2x2

22x = 25

2x=5

x=5/2

Therefore,

log432 =5/2

4. Find x if log5(x-7)=1.

Solution: Given,

log5(x-7)=1

Using logarithm rules, we can write;

51 = x-7

5 = x-7

x=5+7

x=12

5. If logam=n, express an-1 in terms of a and m.

Solution:

logam=n

an=m

an/a=m/a

an-1=m/a

6. Solve for x if log(x-1)+log(x+1)=log21

Solution: log(x-1)+log(x+1)=log21

log(x-1)+log(x+1)=0

log[(x-1)(x+1)]=0

Since, log 1 = 0

(x-1)(x+1) = 1

x2-1=1

x2=2

x=± √2

Since, log of negative number is not defined.

Therefore, x=√2

7. Express log(75/16)-2log(5/9)+log(32/243) in terms of log 2 and log 3.

Solution: log(75/16)-2log(5/9)+log(32/243)

Since, nlogam=logamn

⇒log(75/16)-log(5/9)2+log(32/243)

⇒log(75/16)-log(25/81)+log(32/243)

Since, logam-logan=loga(m/n)

⇒log[(75/16)÷(25/81)]+log(32/243)

⇒log[(75/16)×(81/25)]+log(32/243)

⇒log(243/16)+log(32/243)

Since, logam+logan=logamn

⇒log(32/16)

⇒log2

8. Express 2logx+3logy=log a in logarithm free form.

Solution: 2logx+3logy=log a

logx2+logy3=log a   [By logarithm rule: logab = b log a]

log(x2y3)=log a  [By logarithm rule: log a + log b = log (ab) ]

x2y3 = a  [If logma = logmb, then a = b]