Question Number 79766 by jagoll last updated on 28/Jan/20
$$\mathrm{what}\:\mathrm{is}\:\mathrm{x} \\ $$$$\mathrm{2}^{\mathrm{x}} =\left(\mathrm{1}+\mathrm{tan}\:\mathrm{0}.\mathrm{01}^{\mathrm{o}} \right)\left(\mathrm{1}+\mathrm{tan}\:\mathrm{0}.\mathrm{02}^{\mathrm{o}} \right) \\ $$$$\left(\mathrm{1}+\mathrm{tan}\:\mathrm{0}.\mathrm{03}^{\mathrm{o}} \right)...\left(\mathrm{1}+\mathrm{tan}\:\mathrm{44}.\mathrm{99}^{\mathrm{o}} \right) \\ $$
Answered by mr W last updated on 28/Jan/20
![let′s look at first (1+tan t°)[1+tan (45−t)°] =(1+tan t°)[1+((1−tan t°)/(1+tan t°))] =(1+tan t°)[(2/(1+tan t°))] =2 (1+tan 00.1°)(1+tan 0.02°)...(1+tan 44.99°) =Π_(k=1) ^(4499) (1+tan (k/(100))) ={Π_(k=1) ^(2249) (1+tan (k/(100)))[1+tan (45−(k/(100)))]}(1+tan ((2250)/(100))) =2^(2249) (1+tan ((45)/2)) =2^(2249) (1+(√2)−1) =2^(2249) ×(√2) =2^(2249.5) =2^x ⇒x=2249.5=((4499)/2)](Q79772.png)
$${let}'{s}\:{look}\:{at}\:{first} \\ $$$$\left(\mathrm{1}+\mathrm{tan}\:{t}°\right)\left[\mathrm{1}+\mathrm{tan}\:\left(\mathrm{45}−{t}\right)°\right] \\ $$$$=\left(\mathrm{1}+\mathrm{tan}\:{t}°\right)\left[\mathrm{1}+\frac{\mathrm{1}−\mathrm{tan}\:{t}°}{\mathrm{1}+\mathrm{tan}\:{t}°}\right] \\ $$$$=\left(\mathrm{1}+\mathrm{tan}\:{t}°\right)\left[\frac{\mathrm{2}}{\mathrm{1}+\mathrm{tan}\:{t}°}\right] \\ $$$$=\mathrm{2} \\ $$$$ \\ $$$$\left(\mathrm{1}+\mathrm{tan}\:\mathrm{00}.\mathrm{1}°\right)\left(\mathrm{1}+\mathrm{tan}\:\mathrm{0}.\mathrm{02}°\right)...\left(\mathrm{1}+\mathrm{tan}\:\mathrm{44}.\mathrm{99}°\right) \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{\mathrm{4499}} {\prod}}\left(\mathrm{1}+\mathrm{tan}\:\frac{{k}}{\mathrm{100}}\right) \\ $$$$=\left\{\underset{{k}=\mathrm{1}} {\overset{\mathrm{2249}} {\prod}}\left(\mathrm{1}+\mathrm{tan}\:\frac{{k}}{\mathrm{100}}\right)\left[\mathrm{1}+\mathrm{tan}\:\left(\mathrm{45}−\frac{{k}}{\mathrm{100}}\right)\right]\right\}\left(\mathrm{1}+\mathrm{tan}\:\frac{\mathrm{2250}}{\mathrm{100}}\right) \\ $$$$=\mathrm{2}^{\mathrm{2249}} \left(\mathrm{1}+\mathrm{tan}\:\frac{\mathrm{45}}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}^{\mathrm{2249}} \left(\mathrm{1}+\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$$=\mathrm{2}^{\mathrm{2249}} ×\sqrt{\mathrm{2}} \\ $$$$=\mathrm{2}^{\mathrm{2249}.\mathrm{5}} \\ $$$$=\mathrm{2}^{{x}} \\ $$$$\Rightarrow{x}=\mathrm{2249}.\mathrm{5}=\frac{\mathrm{4499}}{\mathrm{2}} \\ $$
Commented by jagoll last updated on 28/Jan/20
$$\mathrm{sir},\:\mathrm{how}\:\mathrm{many}\:\mathrm{tribes}\:\mathrm{have}\:\mathrm{become} \\ $$$$\mathrm{4499}? \\ $$
Commented by mr W last updated on 28/Jan/20
Commented by mr W last updated on 28/Jan/20
$${is}\:{your}\:{question}\:{answered}\:{now}? \\ $$
Commented by jagoll last updated on 28/Jan/20
$$\mathrm{oo}\:\mathrm{yes}\:\mathrm{sir}\:\mathrm{i}\:\mathrm{understand}\:\mathrm{now}.\:\mathrm{thank}\:\mathrm{you} \\ $$$$\mathrm{sir} \\ $$