Question

if 60^a=3 and 60^b=5 then find the value of (12^1-a-b) ÷2(1-b))

Answer 1

Given :  60ᵃ = 3 ,  60ᵇ = 5

To Find :    $12^{\frac{1-a-b}{2(1-b)}}$

Solution:

60ᵃ = 3 ,  60ᵇ = 5

60ᵃ = 3

=> a log 60 = log 3

=> a = log 3 / log 60

b = log 5 / log 60

(1  - a - b) / (2( 1 - b))

= ( 1  -   log 3 / log 60 -  log 5 / log 60 ) / (2(  1 - log 5 / log 60))

= (  log 60 -  log 3 - log 5) / (2 (log 60 - log 5))

= ( log ( 60/(3*5)))/( 2 log (60/5))

= log 4 / 2 log 12

= 2 log 2 / log 12

= log 2 / log 12

Let say  N  = $12^{\frac{1-a-b}{2(1-b)}}$

substituting (1  - a - b) / (2( 1 - b)) =  log 2 / log 12

=> N =  $12^{ \frac{\log 2}{\log 12} }$

Taking log both sides

log  N  =  log ( $12^{ \frac{\log 2}{\log 12} }$ )

=> log N =   (log 2 / log 12) log 12

=> log N =   log 2

=> N = 2

$12^{\frac{1-a-b}{2(1-b)}}$ = 2

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