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Question Number 213203 by golsendro last updated on 01/Nov/24
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Answered by lepuissantcedricjunior last updated on 01/Nov/24
f(xy)=f(x)+f(y) f(10)=14;f(20)=40 calculons f(500) on a f(500)=f(50×10)=f(50)+f(10) =f(5)+2f(10) or f(20)=f(10)+f(2)=>f(2)=26 f(10)=f(5)+f(2)=>f(5)=−12 =>f(500)=−12+28=16
$$\boldsymbol{{f}}\left(\boldsymbol{{xy}}\right)=\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)+\boldsymbol{{f}}\left(\boldsymbol{{y}}\right) \\ $$$$\boldsymbol{{f}}\left(\mathrm{10}\right)=\mathrm{14};\boldsymbol{{f}}\left(\mathrm{20}\right)=\mathrm{40} \\ $$$$\boldsymbol{{calculons}}\:\boldsymbol{{f}}\left(\mathrm{500}\right) \\ $$$$\boldsymbol{{on}}\:\boldsymbol{{a}}\:\boldsymbol{{f}}\left(\mathrm{500}\right)=\boldsymbol{{f}}\left(\mathrm{50}×\mathrm{10}\right)=\boldsymbol{{f}}\left(\mathrm{50}\right)+\boldsymbol{{f}}\left(\mathrm{10}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\boldsymbol{{f}}\left(\mathrm{5}\right)+\mathrm{2}\boldsymbol{{f}}\left(\mathrm{10}\right) \\ $$$$\boldsymbol{{or}}\:\:\:\boldsymbol{{f}}\left(\mathrm{20}\right)=\boldsymbol{{f}}\left(\mathrm{10}\right)+\boldsymbol{{f}}\left(\mathrm{2}\right)=>\boldsymbol{{f}}\left(\mathrm{2}\right)=\mathrm{26} \\ $$$$\boldsymbol{{f}}\left(\mathrm{10}\right)=\boldsymbol{{f}}\left(\mathrm{5}\right)+\boldsymbol{{f}}\left(\mathrm{2}\right)=>\boldsymbol{{f}}\left(\mathrm{5}\right)=−\mathrm{12} \\ $$$$=>\boldsymbol{{f}}\left(\mathrm{500}\right)=−\mathrm{12}+\mathrm{28}=\mathrm{16} \\ $$
Answered by mr W last updated on 01/Nov/24
f(20)=f(2×10)=f(2)+f(10) ⇒f(2)=40−14=26 f(x^n )=nf(x) f(500)=f(((1000)/2))=f(10^3 ×2^(−1) ) =3f(10)−f(2) =3×14−26=16 ✓
$${f}\left(\mathrm{20}\right)={f}\left(\mathrm{2}×\mathrm{10}\right)={f}\left(\mathrm{2}\right)+{f}\left(\mathrm{10}\right) \\ $$$$\Rightarrow{f}\left(\mathrm{2}\right)=\mathrm{40}−\mathrm{14}=\mathrm{26} \\ $$$${f}\left({x}^{{n}} \right)={nf}\left({x}\right) \\ $$$${f}\left(\mathrm{500}\right)={f}\left(\frac{\mathrm{1000}}{\mathrm{2}}\right)={f}\left(\mathrm{10}^{\mathrm{3}} ×\mathrm{2}^{−\mathrm{1}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3}{f}\left(\mathrm{10}\right)−{f}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3}×\mathrm{14}−\mathrm{26}=\mathrm{16}\:\checkmark \\ $$
Answered by cherokeesay last updated on 01/Nov/24
Answered by Rasheed.Sindhi last updated on 02/Nov/24
f(500)=f(2^2 ×5^3 )=2f(2)+3f(5)...A So we need f(2) and f(5) for f(500) f(10)=f(2×5)=f(2)+f(5)=14...(i) f(20)=f(2^2 ×5)=2f(2)+f(5)=40...(ii) (ii)−(i):f(2)=26 (i)⇒f(5)=14−26=−12 A⇒f(500)=2(26)+3(−12)=16
$${f}\left(\mathrm{500}\right)={f}\left(\mathrm{2}^{\mathrm{2}} ×\mathrm{5}^{\mathrm{3}} \right)=\mathrm{2}{f}\left(\mathrm{2}\right)+\mathrm{3}{f}\left(\mathrm{5}\right)…{A} \\ $$$${So}\:{we}\:{need}\:{f}\left(\mathrm{2}\right)\:{and}\:{f}\left(\mathrm{5}\right)\:{for}\:{f}\left(\mathrm{500}\right) \\ $$$$\: \\ $$$${f}\left(\mathrm{10}\right)={f}\left(\mathrm{2}×\mathrm{5}\right)={f}\left(\mathrm{2}\right)+{f}\left(\mathrm{5}\right)=\mathrm{14}…\left({i}\right) \\ $$$${f}\left(\mathrm{20}\right)={f}\left(\mathrm{2}^{\mathrm{2}} ×\mathrm{5}\right)=\mathrm{2}{f}\left(\mathrm{2}\right)+{f}\left(\mathrm{5}\right)=\mathrm{40}…\left({ii}\right) \\ $$$$\:\left({ii}\right)−\left({i}\right):{f}\left(\mathrm{2}\right)=\mathrm{26} \\ $$$$\left({i}\right)\Rightarrow{f}\left(\mathrm{5}\right)=\mathrm{14}−\mathrm{26}=−\mathrm{12} \\ $$$${A}\Rightarrow{f}\left(\mathrm{500}\right)=\mathrm{2}\left(\mathrm{26}\right)+\mathrm{3}\left(−\mathrm{12}\right)=\mathrm{16} \\ $$

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