Question

30 Points !!!
Do yourself - 1 :
Full Exercise.

Answer 1

1.(a ) 81 = 3⁴
log81 = 3log4

(b) 0.001 = 10-³

log(0.001) = -3log10

(c) 2 = 128^1/7

log2 = 1/7 log128

2) (a) log( 32 base 2) = 5

we know ,
log ( a base b ) = N
then,
a = b^N

use here

so, log( 32 base 2) = 5
32 = 2^5

(b) log{ 4 base √2 } = 4
4 = √2⁴

(c) log( 0.01 base 10 ) = -2
0.01 = 10-²

111) log { 1728 base 2√3 } = x

factors of 1728 = 2× 2× 2× 6 × 6 × 6

1728 = 12³

1728 = (2√3)^6

use this in log

log { (2√3)^6 base 2√3 } = x

we know,

log ( aⁿ base a ) = n

so,

log { (2√3)^6 base (2√3) } = 6 = x

so, x = 6

Answer 2

we know that 
if  N = a^x ⇔ ㏒(base a) N = x

i)
a) 81=3^4 ⇒ log (base 3 ) 81 = 4
b)0.001 =(10)^-1 ⇒ log (base 10) 0.001 = -1
c)2= (128)^1/7 ⇒ log (base 128) 2 = 1/7

ii)
a) log (base 2) 32 = 5 ⇒ 32 = 2^5
b) log (base √2) 4 =4 ⇒ 4 = (√2)^4

c) log (base 10) 0.01 = -2 ⇒ 0.01 = (10)^-1

iii)
1728 = 2^6 * 3³ 
         = 2^6 * (√3)^6
         = (2√3)^6   
now given

log (base 2√3) 1728 =x  ⇒ 1728 = (2√3) ^x

⇒ (2√3) ^6 = (2√3)^x
∴6=x     [ ∵ if a^m = a^n ⇔ m=n  ]
x=6