Question

If log x base 2 + log x base 4 + log x base 16 = 21/4, then x is equal to
(a) 8
(b) 4
(c) 16​

Answer 1

log₂ (x) + log₄ (x) + log₁₆ (x) = 21/4

[ ( log x )/( log 2 ) ] + [ ( log x )/( log 4 ) ] + [ ( log x )/( log 16 ) ] = 21/4

[ ( log x )/( log 2 ) ] + [ ( log x )/( log 2² ) ] + [ ( log x )/( log 2⁴ ) ] = 21/4

[ ( log x )/( log 2 ) ] + [ ( log x )/( 2 log 2 ) ] + [ ( log x )/( 4 log 2 ) ] = 21/4

taking ( log x )/( log 2 ) common

( log x )/( log 2 ) ( 1 + 1/2 + 1/4 ) = 21/4

( log x )/( log 2 ) ( ( 4 + 2 + 2 )/4 ) = 21/4

( log x )/( log 2 ) × 7/4 = 21/4

( log x )/( log 2 ) = 21/4 × 4/7

( log x )/( log 2 ) = 3

log₂ (x) = 3

we can write it as

x = 2³

x = 8

Properties used :

logₐ ( b ) = log b / log a

log aᵇ = b × log a

logₐ ( b ) = y, then b = aʸ