|
5 | added 27 characters in body | ||
|
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, if $f'(x)=-a^{-x}\log a$ 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, if $f'(x)=-a^{-x}\log a$ and $f(x+k)=-a^{-x-k}$ then $k=-logloga/loga$k=-\log \log a/\log a$ |
||||
|
|
3 | added 12 characters in body | ||
|
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 | ||
|
|
1 |
|
||
