Solving for n: 0.99 = 1 - (1 - 0.8)^n

[Shujin]

I've come across this problem in my Prep Course for taking a CQE exam. I've tried different items to try and solve it and have come up empty. The equation asks to solve n. I know n = 2.86 but I'm confused on how to get there. I know I have to use logarithms to solve it, just not sure of the order. The equation follows;

.99=1-(1-.8)^n
.99=1-(.2)^n

Log of .99 = -.00436
Log of .2 = -.6989

This is where I stumble and totally lose my focus. Any help would be greatly appreciated.

Thanks.

 

[pka]

\(\displaystyle \begin{array}{l}
0.99 = 1 - \left( {1 - 0.8} \right)^n \\
0.99 = 1 - \left( {0.2} \right)^n \\
\left( {0.2} \right)^n = 0.01 \\
n = \frac{{\log (0.01)}}{{\log (0.2)}} \\
\end{array}.\)

 

[tkhunny]

Were you given those logarithm values? That seems a little odd, since you may or may not need them.

0.99 = 1 - (0.2)^n

(0.2)^n = 1 - 0.99

(0.2)^n = 0.01

log(0.2^n) = log(0.01)

n*log(0.2) = log(0.01)

n*log(0.2) = -2

n = -2/log(0.2) = -2/(-0.69897) = 2.861353118

I guess I'd have to agree with the offered solution, but I still don't see the point of ALL the given information.

 

[Shujin]

Thanks pka and tkhunny, I really appreciated. Our Engineer from work was helping me with this and that's where I got the log values. Needless to say, we didn't get the answer either.

Sam

 

[Denis]

Sam, just remember that:
if a^p = x
then p = log(x) / log(a)