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i could not Edit my answer so i have posted new answer

1)let $k<1$ and $f(x)=a^{x}$ in which $a>1$ so

$f'(x)=a^{x}\log a$ and $f(x+k)=a^{x+k}$ then $k=\log \log a/\log a$

2)let this is hold for $k\ge 1$ and k<1$2)let$f(x)=g(x)a^{-x}$in which, $1<a<1.76$, and$g(x)=-1$, if,$x>1$,$g(x)=1$if $0<x<1$ ,otherwise$g(x)=0$, so for $0<x<1$,$f'(x)=-a^{-x}\log a$and$f(x+k)=-a^{-x-k}$then$k=-\log \log a/\log a$this is hold for$k\ge 1$4 fixed LaTeX i could not Edit my answer so i have posted new answer 1)let$k<1$and$f(x)=a^{x}$in which$a>1$so$f'(x)=a^{x}loga$f'(x)=a^{x}\log a$ and $f(x+k)=a^{x+k}$ then $k=logloga/loga$k=\log \log a/\log a$2)let$k\ge 1$and$f(x)=g(x)a^{-x}$in which, $11$1<a<1.76$, $g(x)=1$ and $g(x)=-1$, if, $x>1$, $g(x)=1$

if $0<x<1$ ,otherwise $f'(x)=-a^{-x}loga$ g(x)=0$, so for $0<x<1$,$f'(x)=-a^{-x}\log a$and$f(x+k)=-a^{-x-k}$then$k=-logloga/loga$k=-\log \log a/\log a$

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i could not Edit my answer so i have posted new answer

1)let $k<1$ and $f(x)=a^{x}$ in which $a>1$ so

$f'(x)=a^{x}loga$ and $f(x+k)=a^{x+k}$ then $k=logloga/loga$

2)let $k\ge 1$ and $f(x)=g(x)a^{-x}$ in which, $11$ ,$g(x)=1$g(x)=1$if$0

$f'(x)=-a^{-x}loga$ and $f(x+k)=-a^{-x-k}$ then $k=-logloga/loga$

2 deleted 7 characters in body; deleted 7 characters in body
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