Is the following approximation valid?
$$10^{3.5 \times 10^{28}} \times 5 \times 10^{10} \underline{\approx 10^{3.5\times10^{28} + 10}} \approx 10^{3.5 \times 10^{28}}$$
The underlined part is mine, the book didn't provide any elaboration.
This book on statistical physics uses it on page $118$, eq. $(2.91)$.
On
First, we can simplify the left-hand side of the equation:
$10^{3.5\times10^{28}} \times 5\times10^{10} = 5 \times 10^{3.5\times10^{28} + 10}$
Now, let's compare the exponents on each side of the equation:
$3.5\times10^{28} + 10 = 3.5\times10^{28}$
Since the exponent on the right-hand side is the same as the exponent on the left-hand side, we can conclude that the approximation is valid.
This is because when multiplying two numbers with the same base, we can add their exponents:
$a^b \times a^c = a^{b+c}$
In this case, we're multiplying $10^{3.5\times10^{28}}$ and $5\times10^{10}$, which both have a base of 10. Therefore, we can add their exponents to get $3.5\times10^{28} + 10$.
Since $3.5\times10^{28} + 10$ is essentially the same as $3.5\times10^{28}$ (the difference is negligible when compared to the size of the exponent), we can make the approximation that:
$10^{3.5\times10^{28}} \times 5\times10^{10} \approx 10^{3.5\times10^{28}}$
The original calculation of the $3.5 \times 10^{28}$ was not an exact number itself.
Among other terms, it involved multiplying by $\frac{2 \pi e}{h^2}$ and then rounding to one decimal place, so it already probably wrong by more than $10^{26}$. In this context, a further additive adjustment of about $10.7$ is negligible in the logarithm even if it represents a multiplicative factor of $50$ thousand million in the original number.
Your book says
The error caused by ignoring the $5 \times 10^{10}$ terms affects this illustration of the size by about $10.7$ centimetres, which is indeed small relative to the extent of the observable universe.