SL 1.5—Intro to logs

In $7^3=343$, the base is $7$, the exponent is $3$, and the result is $343$.

The rule states: $a^b=c⟺{\log }_{a}c=b$. Here $a=7$, $b=3$, $c=343$, so the logarithmic form is ${\log }_{7}343=3$.

The base of the log is $7$, the argument is $343$, and the result is $3$. This matches ${\log }_{7}343=3$.

We need the logarithmic form of $7^3=343$. The base of the logarithm must be $7$, since $7$ is the base of the exponent.

${\log }_{3}343=7$ has base $3$, not $7$ — this is incorrect. ${\log }_{343}7=3$ has base $343$, which is the result, not the base — eliminated.

${\log }_{7}3=343$ claims the log equals $343$, but the exponent is $3$, not $343$ — this is clearly wrong.

Only ${\log }_{7}343=3$ correctly places $7$ as the base, $343$ as the argument, and $3$ as the result.

Let ${\log }_{4}64=b$. This means $4^b=64$.

$4^1=4$, $4^2=16$, $4^3=64$. So $b=3$.

${\log }_{4}64=3$.

We need the power to which $4$ must be raised to get $64$.

$4^2=16\ne 64$, so the answer is not $2$.

$4^4=256\ne 64$, so $4$ is wrong. $16$ is far too large to be an exponent here — $4^{16}$ is enormous.

$4^3=64$ ✓, so ${\log }_{4}64=3$.