Menu Close

Question-157769




Question Number 157769 by Study last updated on 27/Oct/21
Commented by Study last updated on 27/Oct/21
what is the practice and solve?
$${what}\:{is}\:{the}\:{practice}\:{and}\:{solve}? \\ $$
Commented by mr W last updated on 27/Oct/21
1.1^n =2  n=((log 2)/(log 1.1))≈7.3 years
$$\mathrm{1}.\mathrm{1}^{{n}} =\mathrm{2} \\ $$$${n}=\frac{\mathrm{log}\:\mathrm{2}}{\mathrm{log}\:\mathrm{1}.\mathrm{1}}\approx\mathrm{7}.\mathrm{3}\:{years} \\ $$
Commented by Study last updated on 28/Oct/21
please helpe my[
$${please}\:{helpe}\:{my}\left[\right. \\ $$
Commented by Study last updated on 28/Oct/21
can we solve it by simple intersect formula?
$${can}\:{we}\:{solve}\:{it}\:{by}\:{simple}\:{intersect}\:{formula}? \\ $$
Commented by mr W last updated on 28/Oct/21
P(1+((10)/(100)))^n =2P  1.1^n =2  ⇒n=((log 2)/(log 1.1))  that′s all.
$${P}\left(\mathrm{1}+\frac{\mathrm{10}}{\mathrm{100}}\right)^{{n}} =\mathrm{2}{P} \\ $$$$\mathrm{1}.\mathrm{1}^{{n}} =\mathrm{2} \\ $$$$\Rightarrow{n}=\frac{\mathrm{log}\:\mathrm{2}}{\mathrm{log}\:\mathrm{1}.\mathrm{1}} \\ $$$${that}'{s}\:{all}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *