Question

if log base 2 x +log base 4x +log base 16 x=21/4 ,then x =​

Answer 1

Answer:

We know that log

a

n

N=

n

1

log

a

N.

Now,

log

2

x+log

4

x+log

16

x=

4

21

⇒log

2

x+log

2

2

x+log

2

4

x=

4

21

⇒log

2

x+

2

1

log

2

x+

4

1

log

2

x=

4

21

,

⇒(1+

2

1

+

4

1

)log

2

x=

4

21

⇒

4

7

log

2

x=

4

21

⇒log

2

x=3

Hence, x=2

3

=8

Answer 2

Answer:

$log_{2}(x) + log_{4}(x) + log_{16}(x) = \frac{21}{4}$

$log_{2}(x) + log_{ {2}^{2} }(x) + log_{ {2}^{4} }(x) = \frac{21}{4}$

$log_{2}(x) + log_{2}( {x}^{ \frac{1}{2} } ) + log_{2}( {x}^{ \frac{1}{4} } ) = \frac{21}{4}$

$log_{2}( {x}^{1 + \frac{1}{2} + \frac{1}{4} } ) = \frac{21}{4}$

${x}^{ \frac{7}{4} } = {2}^{ \frac{21}{4} }$

${x}^{ \frac{7}{2} } = {8}^{ \frac{7}{2} }$

$x = 8$