Question

If log2x + log4x + log16x = 21/4, these x is equal to​

Answer 1

we know that:

log a + log b = log(ab)

then ,

$log_{}(2x) + log(4x) + log(16x)$

$= log(128x {}^{3} )$

this implies

$log(128x {}^{3} ) = \frac{21}{4}$

if base of log is 10

then ,

$10 {}^{ \frac{21}{4} } = 128x {}^{3}$

now solve the expression

you get the answer

Answer 2

Answer:

Step-by-step explanation: