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**Prove It** For starters, you need to remember the logarithm law $\displaystyle \displaystyle \begin{align*} \log_b{ \left( m^p \right) } = p\log_b{(m)} \end{align*}$. So if you start with your equation $\displaystyle \displaystyle \begin{align*} 3^{4567} = 10^x \end{align*}$ and take a logarithm of ANY base to both sides (in this case they chose 3 because it cancels easiest on the left hand side), we have

$\displaystyle \displaystyle \begin{align*} 3^{4567} &= 10^x \\ \log_3{\left( 3^{4567} \right) } &= \log_3{\left( 10^x \right) } \\ 4567 &= \log_3{ \left( 10^x \right) } \\ 4567 &= x \log_3{(10)} \textrm{ from the logarithm law mentioned earlier} \\ \frac{4567}{\log_3{(10)}} &= x \end{align*}$

It would have been just as easy to cancel on the RHS by taking both sides to a base 10 logarithm, or even taking both sides to any other logarithm as well.

Most logarithm values are transcendental, unless the number you are taking a logarithm of is a power of the same base of your logarithm, because then you have natural cancellation. If your logarithm is transcendental though, decimal approximations can be made using a Taylor Series (which calculators would be programmed with).