Question 798721
With each reduction, the previous dimensions get multiplied by {{{0.64=64/100="64%"}}}.
With {{{n}}} reductions the original dimensions will be multiplied times {{{0.64^n}}},
and they will end up being multiplied times {{{"0.10"=10/100="10 %"}}}.
So {{{0.64^n<0.10}}} is our equation

We could say that thew original length would be {{{b[0]}}},
and the length of the nth reduction would be {{{b[n]=b[0]*0.64^n}}}.
Then we could write that a 10% reduction would reduce {{{b[0]}}} to {{{b[0]*0.1}}},
and write {{{b[0]*0.64^n<b[0]*0.1}}} --> {{{0.64^n<0.10}}}
 
From there we can
either start calculating powers of 0.64 until we get to less than 0.1,
or use logarithms.
 
Calculating powers:
{{{0.64^2=0.4096}}}
{{{0.64^3=0.262144}}}
{{{0.64^4=0.167772}}}
{{{0.64^5=0.107374}}}
{{{0.64^6=0.0687194}}}
 
Using logarithms:
{{{0.64^n<0.10}}} --> {{{log(0.64^n)<log(0.10)}}} --> {{{n*log(0.64)<-1}}} --> {{{n*log(0.64)<-1}}}
{{{log(0.64)=approximately}}}{{{-0.19382<0}}}
Dividing both sides of the inequality by a negative number, the inequality sign reverses, so
{{{n*log(0.64)<-1}}} --> {{{n>-1/log(0.64)=approximately}}}{{{5.16}}}
Since we need an integer {{{n}}}, we need {{{n>=6}}}