Editing List of logarithmic identities (section)

== Cancelling exponentials ==
Logarithms and [[Exponential function|exponentials]] with the same base cancel each other. This is true because logarithms and exponentials are inverse operations—much like the same way multiplication and division are inverse operations, and addition and subtraction are inverse operations.

:<math>b^{\log_b(x)} = x\text{ because }\mbox{antilog}_b(\log_b(x)) = x</math>

:<math>\log_b(b^x) = x\text{ because }\log_b(\mbox{antilog}_b(x)) = x</math><ref>{{Cite web|last=Weisstein|first=Eric W.|title=Logarithm|url=https://mathworld.wolfram.com/Logarithm.html|access-date=2020-08-29|website=mathworld.wolfram.com|language=en}}</ref>

Both of the above are derived from the following two equations that define a logarithm:
(note that in this explanation, the variables of <math> x</math> and <math>x</math> may not be referring to the same number)

:<math>\log_b (y) = x  \iff  b^x = y</math>

Looking at the equation <math> b^x = y </math>, and substituting the value for <math>x</math> of
<math> \log_b (y) = x </math>, we get the following equation: 
<math>
b^x = y
\iff b^{\log _b(y)} = y
\iff b^{\log_b (y)} = y
</math>
, which gets us the first equation.
Another more rough way to think about it is that <math> b^{\text{something}} = y</math>,
and that that "<math>\text{something}</math>" is <math> \log_b (y) </math>.

Looking at the equation <math> \log_b (y) = x</math>
, and substituting the value for <math> y </math> of <math>b^x = y</math>, we get the following equation:
<math>
\log_b (y) = x
\iff \log_b(b^x) = x
\iff \log_b(b^x) = x
</math>
, which gets us the second equation.
Another more rough way to think about it is that <math> \log_b (\text{something}) = x</math>,
and that that something "<math>\text{something}</math>" is <math> b^x</math>.
<!---   previously what was there:
:<math>b^c = x \iff \log_b(x) = c</math>

Substituting {{mvar|c}} in the left equation gives {{math|1=''b''<sup>log<sub>''b''</sub>(''x'')</sup> = ''x''}}, and substituting {{mvar|x}} in the right gives {{math|1=log<sub>''b''</sub>(''b''<sup>''c''</sup>) = ''c''}}. Finally, replace {{mvar|c}} with {{mvar|x}}.
-->