fortressinthesky
fortressinthesky fortressinthesky
  • 19-08-2021
  • Mathematics
contestada

If [tex]f(x)=3^x[/tex], prove that [tex]f(x) + f(x+1)=4f(x)[/tex]

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WojtekR
WojtekR WojtekR
  • 19-08-2021

[tex]$f(x)+f(x+1) \stackrel{1}{=} 3^x+3^{x+1} \stackrel{2}{=} 3^x+3^x\cdot3^1\stackrel{3}{=}3^x+3\cdot3^x\stackrel{4}{=}4\cdot3^x\stackrel{5}{=}4\cdot f(x)[/tex]

1) Substitution from definition.

2) We use [tex]$x^{a+b}=x^a\cdot x^b[/tex]

3) We are changing the order of 3 and [tex]3^x[/tex].

4) We add 1 and 3 [tex]3^x[/tex] and we get [tex]4\cdot 3^x[/tex]

5) Substitution from definition (but the other way than at the beginning).

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