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Question Number 118118 by naka3546 last updated on 15/Oct/20

f(x+y) = f(x) f(y)   ∀x ∈ R  f(1) = 8  f((2/3)) = ?

$${f}\left({x}+{y}\right)\:=\:{f}\left({x}\right)\:{f}\left({y}\right)\:\:\:\forall{x}\:\in\:\mathbb{R} \\ $$$${f}\left(\mathrm{1}\right)\:=\:\mathrm{8} \\ $$$${f}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\:=\:? \\ $$

Answered by mr W last updated on 15/Oct/20

f(x)=a^x   f(1)=a^1 =8 ⇒a=8  f(x)=8^x   f((2/3))=8^(2/3) =4

$${f}\left({x}\right)={a}^{{x}} \\ $$$${f}\left(\mathrm{1}\right)={a}^{\mathrm{1}} =\mathrm{8}\:\Rightarrow{a}=\mathrm{8} \\ $$$${f}\left({x}\right)=\mathrm{8}^{{x}} \\ $$$${f}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\mathrm{8}^{\frac{\mathrm{2}}{\mathrm{3}}} =\mathrm{4} \\ $$

Commented by bemath last updated on 15/Oct/20

sir, how do you claim that f(x)=a^x  ?

$${sir},\:{how}\:{do}\:{you}\:{claim}\:{that}\:{f}\left({x}\right)={a}^{{x}} \:? \\ $$

Commented by MJS_new last updated on 15/Oct/20

a^(x+y) =a^x a^y

$${a}^{{x}+{y}} ={a}^{{x}} {a}^{{y}} \\ $$

Commented by bemath last updated on 15/Oct/20

okay prof thank you

$${okay}\:{prof}\:{thank}\:{you} \\ $$

Answered by mindispower last updated on 15/Oct/20

f((2/3))=f((1/3))^2   f((2/3)+(1/3))=f(1)=f((2/3)).f((1/3))=f((1/3))^3 =8⇒f((1/3))=2  f((2/3))=2^2 =4

$${f}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)={f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$${f}\left(\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\right)={f}\left(\mathrm{1}\right)={f}\left(\frac{\mathrm{2}}{\mathrm{3}}\right).{f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)={f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} =\mathrm{8}\Rightarrow{f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{2} \\ $$$${f}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\mathrm{2}^{\mathrm{2}} =\mathrm{4} \\ $$$$ \\ $$

Answered by aleks041103 last updated on 15/Oct/20

f(x+y)=f(x)f(y)  ln(f(x+y))=ln(f(x)f(y))  ln(f(x+y))=ln(f(x)) + ln(f(y))  g(x)=ln(f(x))  ⇒g(x+y)=g(x)+g(y)  let y=0 ⇒ g(x)=g(x)+g(0)⇒g(0)=0  g(Σ_(i=1) ^n x_i )=g(x_n +Σ_(i=1) ^(n−1) x_i )=g(x_n )+g(Σ_(i=1) ^(n−1) x_i )  by induction  g(Σ_(i=1) ^n x_i )=Σ_(i=1) ^n g(x_i )  if x_i =x  g(nx)=ng(x), n is integer  ln(f(nx))=n ln(f(x))=ln((f(x))^n )  ⇒f(nx)=(f(x))^n   ⇒f(3×(1/3))=(f((1/3)))^3 = 8  ⇒f(1/3)=8^(1/3)   f(2×(1/3))=(f((1/3)))^2 =8^(2/3)   ⇒f((2/3))=4 1^(2/3) =4(e^(2kπi) )^(2/3)   ⇒f((2/3))=4exp(((4k)/3)πi)=4cos(240°k)+4sin(240°k)i  ⇒f(2/3)=4;−2+2(√3)i;−2−2(√3)i

$${f}\left({x}+{y}\right)={f}\left({x}\right){f}\left({y}\right) \\ $$$${ln}\left({f}\left({x}+{y}\right)\right)={ln}\left({f}\left({x}\right){f}\left({y}\right)\right) \\ $$$${ln}\left({f}\left({x}+{y}\right)\right)={ln}\left({f}\left({x}\right)\right)\:+\:{ln}\left({f}\left({y}\right)\right) \\ $$$${g}\left({x}\right)={ln}\left({f}\left({x}\right)\right) \\ $$$$\Rightarrow{g}\left({x}+{y}\right)={g}\left({x}\right)+{g}\left({y}\right) \\ $$$${let}\:{y}=\mathrm{0}\:\Rightarrow\:{g}\left({x}\right)={g}\left({x}\right)+{g}\left(\mathrm{0}\right)\Rightarrow{g}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${g}\left(\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{x}_{{i}} \right)={g}\left({x}_{{n}} +\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{x}_{{i}} \right)={g}\left({x}_{{n}} \right)+{g}\left(\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{x}_{{i}} \right) \\ $$$${by}\:{induction} \\ $$$${g}\left(\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{x}_{{i}} \right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{g}\left({x}_{{i}} \right) \\ $$$${if}\:{x}_{{i}} ={x} \\ $$$${g}\left({nx}\right)={ng}\left({x}\right),\:{n}\:{is}\:{integer} \\ $$$${ln}\left({f}\left({nx}\right)\right)={n}\:{ln}\left({f}\left({x}\right)\right)={ln}\left(\left({f}\left({x}\right)\right)^{{n}} \right) \\ $$$$\Rightarrow{f}\left({nx}\right)=\left({f}\left({x}\right)\right)^{{n}} \\ $$$$\Rightarrow{f}\left(\mathrm{3}×\frac{\mathrm{1}}{\mathrm{3}}\right)=\left({f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right)^{\mathrm{3}} =\:\mathrm{8} \\ $$$$\Rightarrow{f}\left(\mathrm{1}/\mathrm{3}\right)=\mathrm{8}^{\mathrm{1}/\mathrm{3}} \\ $$$${f}\left(\mathrm{2}×\frac{\mathrm{1}}{\mathrm{3}}\right)=\left({f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right)^{\mathrm{2}} =\mathrm{8}^{\mathrm{2}/\mathrm{3}} \\ $$$$\Rightarrow{f}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\mathrm{4}\:\mathrm{1}^{\mathrm{2}/\mathrm{3}} =\mathrm{4}\left({e}^{\mathrm{2}{k}\pi{i}} \right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\Rightarrow{f}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\mathrm{4}{exp}\left(\frac{\mathrm{4}{k}}{\mathrm{3}}\pi{i}\right)=\mathrm{4}{cos}\left(\mathrm{240}°{k}\right)+\mathrm{4}{sin}\left(\mathrm{240}°{k}\right){i} \\ $$$$\Rightarrow{f}\left(\mathrm{2}/\mathrm{3}\right)=\mathrm{4};−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}{i};−\mathrm{2}−\mathrm{2}\sqrt{\mathrm{3}}{i} \\ $$

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