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Algebra Level 2

Find the smallest positive integer $x$ such that $(1 + 0.0001)^{x}$ is greater than $2$ .

Details and Assumptions :

1 solution

Milind Prabhu
Jan 21, 2015

${ \left( 1.0001 \right) }^{ x }>2$

$xlog(1.0001)>log2$

$x>\frac { log2 }{ log(1.0001) }$

Since we are looking for the smallest such integer,

$x=\left\lceil \frac { log2 }{ log(1.0001) } \right\rceil$

$x=\left\lceil 6931.81837341 \right\rceil$

$x=\boxed { 6932 }$