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Question

(C) Assume that $\log _{5} 2$ is a rational number,so $\log _{5} 2 = \frac{p}{q}$,where $p$ and $q$ are integers and $q 
eq 0$.By the definition of l

Vedclass · Accepted Answer

an irrational number

Vedclass · Answer

(C) Assume that $\log _{5} 2$ is a rational number,so $\log _{5} 2 = \frac{p}{q}$,where $p$ and $q$ are integers and $q 
eq 0$.By the definition of logarithms,this implies $2 = 5^{p/q}$.Raising both sides to the power of $q$,we get $2^q = 5^p$.Since $2^q$ is an even number (for $q \geq 1$) and $5^p$ is an odd number,this leads to a contradiction.Therefore,the assumption that $\log _{5} 2$ is rational is false.Hence,$\log _{5} 2$ is an irrational number.

$\log _{5} 2$ is

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