DIAMMIBENVERMk1

2022-07-08

Using Logarithms
$\begin{array}{rl}-{2}^{n-1}\mathrm{ln}2& =-100\mathrm{ln}10\\ & \\ -100\mathrm{ln}10& =-230\\ & \\ \frac{-230}{\mathrm{ln}\left(2\right)}& =-333\\ & \\ -{2}^{n-1}& >-333\\ & \\ \left(n-1\right)\mathrm{ln}\left(-2\right)& >\mathrm{ln}\left(-333\right)\end{array}$
Here is where I am stuck.
I am not sure if this part is correct: $-n-1=8$. Then solving we would get $-9$

Camron Herrera

Expert

Log is an increasing function. So you can take log in both side of an inequality and that is fine. But you can't take log of negative values like -2 The mistake is at
$-{2}^{n-1}=-333.$
You missed negative sign at 333. Then if you correct the next steps it's fine. Also when you multiply both sides of an inequality by negative the sign changes.
$-{2}^{n-1}ln2<-100\ast ln10$
$-{2}^{n-1}<\frac{-100\ast ln10}{ln2}$
${2}^{n-1}>\frac{100\ast ln10}{ln2}$
$n-1\ast ln\left(2\right)>ln\left(\frac{100\ast ln10}{ln2}\right).$
And this gives $n>\frac{ln\left(\frac{100\ast ln10}{ln2}\right)}{ln2}+1$

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