SOLUTION: Most photocopiers can reduce the size an image by a maximum of 64% of the original dimensions how many reductions at the maximum setting would it take to reduce an image to less th

Algebra ->  Sequences-and-series -> SOLUTION: Most photocopiers can reduce the size an image by a maximum of 64% of the original dimensions how many reductions at the maximum setting would it take to reduce an image to less th      Log On


   

Question 798721: Most photocopiers can reduce the size an image by a maximum of 64% of the original dimensions how many reductions at the maximum setting would it take to reduce an image to less than 10% of its original dimensions
I did it logically and I got a answer of 6
But I prefer to do it algebraically using the geometric sequence formula
Show me !!

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
With each reduction, the previous dimensions get multiplied by 0.64=64%2F100=%2264%25%22.
With n reductions the original dimensions will be multiplied times 0.64%5En,
and they will end up being multiplied times %220.10%22=10%2F100=%2210+%25%22.
So 0.64%5En%3C0.10 is our equation
We could say that thew original length would be b%5B0%5D,
and the length of the nth reduction would be b%5Bn%5D=b%5B0%5D%2A0.64%5En.
Then we could write that a 10% reduction would reduce b%5B0%5D to b%5B0%5D%2A0.1,
and write b%5B0%5D%2A0.64%5En%3Cb%5B0%5D%2A0.1 --> 0.64%5En%3C0.10

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%5E2=0.4096
0.64%5E3=0.262144
0.64%5E4=0.167772
0.64%5E5=0.107374
0.64%5E6=0.0687194

Using logarithms:
0.64%5En%3C0.10 --> log%280.64%5En%29%3Clog%280.10%29 --> n%2Alog%280.64%29%3C-1 --> n%2Alog%280.64%29%3C-1
log%280.64%29=approximately-0.19382%3C0
Dividing both sides of the inequality by a negative number, the inequality sign reverses, so
n%2Alog%280.64%29%3C-1 --> n%3E-1%2Flog%280.64%29=approximately5.16
Since we need an integer n, we need n%3E=6