Are these exponential functions?

Question

I learnt that exponential functions has the format : \begin{align} number^x \end{align}

But these functions are not exactly power x, so I am confused:
\begin{align} \sqrt{5}^{\log_{5} x}\\ 8^{\lg x} \end{align}
I am trying to understand the classes of function, so I would like to know what class are the above 2 functions in

Answer 1

Note that $$ \sqrt 5 ^{\log_5 x } =5^{1/2 \log_5 x}=\sqrt {x} $$

Which is not exponential.

Also $$8^{\log x} =x^{\log 8}$$

Which is not exponential

Thus none of your expressions is exponential.

Answer 2

In hindsight I think the text where you found these functions wanted you to use the identity:

$a^{\log_c b} = b^{\log_c a}$

(Proof: $(a)^{\log_c b} =$ $(c^{\log_c a})^{\log_c b}=$ $c^{\log_c a\cdot \log_c b}=$ $(c^{\log_c b})^{\log_c a} =$ $(b)^{\log_c a}$.)

So $\sqrt 5^{\log_5 x} = x^{\log_5 \sqrt 5} = x^{\frac 12} = \sqrt x$.

And $8^{\lg 2 x}= x^{\lg 8} = x^3$.

So ....

A function $b^{x}$ is an exponential function. A function $x^b$ is a power function. (They are different.) A function $b^{\log_k x}=x^{\log_k b}$ is a power function in disguise.

====== old answer =====

Okay....

An exponential function is of the form $b^x$. You are describing functions of the form $b^{f(x)}$ which are not exponential functions but clearly related.

In particular you are talking of functions of the form $b^{\log_k x}$.

Now we can change the base of an exponential function by noting that

$b^x = (b)^x = (k^{\log_k b})^x = k^{(x\cdot \log_k b}=k^{dx}$ where $d = \log_k b$. Or by noting $b^x = b^{ax*\frac 1a} = c^{ax} $ where $c = b^{\frac 1a}$.

So exponential functions can be in the form $b^{ax}$ for a constant $a$.

Your functions $f(x) = b^{\log_k x}$ can have a change of base as well.

$(b)^{\log_k x} = (k^{\log_k b})^{\log_k x} = k^{\log_k b\cdot \log_k x} = (k^{\log_k x})^{\log_k b} = x^{\log_k b} = x^c$ where $c = \log_k b$ is some constant.

So ....

Exponential functions are of the form $b^x$. Power functions are of the form $x^c$. Yours are actually examples of power functions.

[My first version overlooked the obvious and made a very mistaken conclusion. You can view the edit history to see a naive, not incorrect but utterly naive, earlier take about your functions.]