$a$, $b$, $c$, $d$ and $x$ are five literals. The product of the terms logarithm of $a$ to $b$, logarithm of $b$ to $c$, logarithm of $c$ to $d$ and logarithm $d$ to $x$ becomes an exponent to the literal number $a$.

$\large a^{\displaystyle \log_{\displaystyle a} b . \log_{\displaystyle b} c . \log_{\displaystyle c} d . \log_{\displaystyle d} x}$

The product of the logarithmic terms follows a sequence and it allows us to simplify the expression easily.

Consider the product of first two logarithmic terms.

$= \large a^{\displaystyle (\log_{\displaystyle a} b . \log_{\displaystyle b} c) . \log_{\displaystyle c} d . \log_{\displaystyle d} x}$

Now use change of base rule of logarithm to combine the product of them as a logarithmic term.

$= \large a^{\displaystyle (\log_{\displaystyle a} c ) . \log_{\displaystyle c} d . \log_{\displaystyle d} x}$

$= \large a^{\displaystyle \log_{\displaystyle a} c . \log_{\displaystyle c} d . \log_{\displaystyle d} x}$

Once again, consider the product of first two logarithm terms and use same change of base log rule to simply the product of them.

$= \large a^{\displaystyle (\log_{\displaystyle a} c . \log_{\displaystyle c} d) . \log_{\displaystyle d} x}$

$= \large a^{\displaystyle (\log_{\displaystyle a} d) . \log_{\displaystyle d} x}$

$= \large a^{\displaystyle \log_{\displaystyle a} d . \log_{\displaystyle d} x}$

Similarly, repeat the same procedure one more time.

$= \large a^{\displaystyle \log_{\displaystyle a} d . \log_{\displaystyle d} x}$

$= \large a^{\displaystyle \log_{\displaystyle a} x}$

There is only one logarithm term as an exponent of the literal a and the base of the logarithm is also same. It can be simplified by the fundamental logarithmic identity.

$= \large a^{\displaystyle \log_{\displaystyle a} x}$

$= \large x$

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