For a reaction, $\frac{{\mathrm{K}}_{\mathrm{t}+10}}{{\mathrm{K}}_{\mathrm{t}}}=\mathrm{x}$ When temperature is increased from 10⁰C to 100⁰C, rate constant
(K) increased by a factor of 512. Then, value of x is

Temperature difference i.e. ΔT=T2​−T1​=100−10=90

$\frac{{\mathrm{K}}_{\mathrm{T}2}}{{\mathrm{K}}_{\mathrm{T}1`}}$=512

$\frac{{\mathrm{K}}_{\mathrm{t}}+10}{{\mathrm{K}}_{\mathrm{t}}}=\mathrm{x}$

To find x ?
 

So,$\frac{{\mathrm{K}}_{\mathrm{T}2}}{{\mathrm{K}}_{\mathrm{T}1`}}$ =${\mathrm{\mu }}^{\frac{∆\mathrm{T}}{10}}$

μ is temperature coefficient of reaction.

On putting all  the given values in this we will find: 

512=${\mathrm{\mu }}^{\frac{∆\mathrm{T}}{10}}$

512=​ ${\mathrm{\mu }}^{\frac{90}{10}}$

$2^9 = {\mathrm{\mu }}^9$

so μ=2

which is x in this question