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$$5\times 1.2^x-3$$

Is it simply that it doesn't follow the form "$ca^x$"?

Siong Thye Goh
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Z.Apa
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3 Answers3

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It doesn't follow that $\,f(x+y)=f(x)f(y)\,$, which is what makes a function eligible to be even considered as being called exponential. For why not all functions that obey the above are actually exponential, see the answers under Overview of basic facts about Cauchy functional equation.

[ EDIT ]  If using a definition of exponentials which includes functions of the form $\,\color{red}{c \cdot} a^x\,$, then the equality above needs to be changed to $\,f(0)f(x+y)=f(x)f(y)\,$.
dxiv
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  • But then ca^x wouldn't be an exponential function either?

    (Sorry about the math, format...)

     – Z.Apa Jan 25 '18 at 09:01
  • @Z.Apa The exponential functon is $,e^x,$, and functions of the form $,e^{\lambda x} = a^x,$ are also called exponentials pretty much everywhere. Those of the form $,e^{\lambda x+\mu} = ca^x,$ may or may not be called exponential functions depending on the definition you are using. I added a clarification at the end of the answer. – dxiv Jan 25 '18 at 18:01
  • I hope you don't mind an exception to the rule against comments like "thanks", but your addition really made a huge difference. So thank you! – Z.Apa Jan 26 '18 at 07:50
  • @Z.Apa Glad it helped. Also prompted me to re-read that addition and correct the "edit" vs. "exit" typo ;-) – dxiv Jan 26 '18 at 08:02
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Yes, the $-3$ is exactly what makes your function not exponential.

Arthur
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Yes, it can't be written in that form. Notice that function of the form of $ca^x$ doesn't change sign.

But for our function $f(x)=5 \times 1.2^x-3$,

$$\lim_{x \to -\infty}f(x)=-3$$ and $$\lim_{x \to \infty}f(x)=\infty.$$

Siong Thye Goh
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