How to solve this equation

[jess5085]

Hello,

I'm not sure how to solve this equation.

128*(1/2)^n-1 = 1/512 for n

Any help would be appreciated.

Thanks!

 

[Ishuda]

Hello,

I'm not sure how to solve this equation.

128*(1/2)^n-1 = 1/512 for n

Any help would be appreciated.

Thanks!

First I'll take the equation as
128*(1/2)^(n-1) = 1/512
and re-write it as
\(\displaystyle (\frac{1}{2})^{n-1} = \frac{1}{128 * 512}\)
The easiest way is to recognize that 4*128=512 and note that 128=2⁷. So 128*512 = 2² 2⁷ 2⁷ = 2¹⁶ and proceed from there.

However, that is not the general way to solve problems of this type unless you carry all powers of numbers in you head. If you have studied logarithms, then you can use them to find the answer. Proceed as before and re-write as
\(\displaystyle (\frac{1}{2})^{n-1} = \frac{1}{65536}\)
You can proceed by taking the logarithm of each side and go from there. You could, if you wish cross multiply to get
\(\displaystyle 65536 = 2^{n-1}\)
first before taking the logarithms.