## Intermediate Algebra (6th Edition)

$\frac{log(7)}{log(2)}$ or $\frac{ln(7)}{ln(2)}$
The change of base formula can be written as $log_{b}a=\frac{log_{c}a}{log_{c}b}$ (where a, b, and c are positive real numbers and neither b nor c is equal to 1). Therefore, any form of $log_{2}7=log_{b}a$ written in another base will be of the form $\frac{log_{c}7}{log_{c}2}$. Answer choice a is in this form, where $c=10$. Answer choice b is also written in this form, where $c=e$. Recall that $ln (7)$ can be written as $log_{e}7$, for example. So, $log_{2}7$ is equivalent to answer choices a and b.