How is log base 10 related to the Natural Logarithm

Question

Log base 4 (7) can be evaluated using the change of base formula. This can similarly be expressed as ln7 divided by ln4, using the natural logarithm.

Using the above example,

Why is it that these two ratios are equal, but log base 10 (7) is not equal to ln 7?

More generally, why can the log base 10 (x) be replaced in the change of base calculation with lnx?

Answer 1

It is true that:

$$\log_a(x) = \dfrac{\log_b(x)}{\log_b(a)}$$ for all positive real values of $b$ different than $1$. In particular when $b=10$ or when $b=e$.

To see this... suppose that $\log_a(x)=y$.

Then you have $x=a^y$.

Now, take the logarithm base $b$ of both sides to get $\log_b(x)=\log_b(a^y)$ which by logarithm rules the exponent can drop outside as $\log_b(x)=y\log_b(a)$

Now, dividing you get $y=\dfrac{\log_b(x)}{\log_b(a)}$ which remembering that $y=\log_a(x)$ gives the identity given at the start.

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If your question really was why $\log_{10} 7 \neq \log_e 7$, that should be obvious since $10\neq e$

Answer 2

The basic idea is that: $$ 10\approx\operatorname{e}^{2.3} $$ and therefore $\operatorname{e}$ has to be raised to approximately $2.3$ times as high a power as does $10$ in order to yield the same result. Hence: $$ \ln(x)\approx2.3\log_{10}(x) $$ and so: $$ \frac{\ln(a)}{\ln(b)}=\frac{2.3\log_{10}(a)}{2.3\log_{10}(b)} $$ and the common factor $\ln(10)\approx2.3$ cancels out.