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Question Number 23922 by chernoaguero@gmail.com last updated on 09/Nov/17

if f(x)=2^x   show that f(x+3)−f(x−1)=((15)/2)f(x)

$$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\mathrm{2}^{\boldsymbol{\mathrm{x}}} \\ $$$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}+\mathrm{3}\right)−\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)=\frac{\mathrm{15}}{\mathrm{2}}\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$

Answered by $@ty@m last updated on 10/Nov/17

f(x+3)=2^(x+3) =2^x .2^3 =8.2^x   f(x−1)=2^(x−1) =2^x .2^(−1) =(2^x /2)  ∴LHS=f(x+3)−f(x−1)              =2^x (8−(1/2))              =((15)/2)f(x)=RHS

$${f}\left({x}+\mathrm{3}\right)=\mathrm{2}^{{x}+\mathrm{3}} =\mathrm{2}^{{x}} .\mathrm{2}^{\mathrm{3}} =\mathrm{8}.\mathrm{2}^{{x}} \\ $$$${f}\left({x}−\mathrm{1}\right)=\mathrm{2}^{{x}−\mathrm{1}} =\mathrm{2}^{{x}} .\mathrm{2}^{−\mathrm{1}} =\frac{\mathrm{2}^{{x}} }{\mathrm{2}} \\ $$$$\therefore{LHS}=\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}+\mathrm{3}\right)−\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}^{{x}} \left(\mathrm{8}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{15}}{\mathrm{2}}\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)={RHS} \\ $$

Commented by chernoaguero@gmail.com last updated on 10/Nov/17

Thank u sir God bless u

$$\mathrm{Thank}\:\mathrm{u}\:\mathrm{sir}\:\mathrm{God}\:\mathrm{bless}\:\mathrm{u} \\ $$

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