Given $\log_m 2 = a$ and $\log_m 3 = b$, I want to calculate $\log_m (64/2.7)- \log_m 60$. Using the logarithm properties, and simplifying, I get $5a + b - \log 27/\log m$, where $\log x$ is the $\log$ in base 10. But it seems it could be more simplified...
It is possible to simplify it more?
2
$\begingroup$
algebra-precalculus
2 Answers
4
we have: $$ \log_m (64/2.7)- \log_m 60 = \log_m\frac{64 \cdot 10}{27 \cdot 60}= $$ $$ =\log_m\frac{2^5}{3^4}=5a-4b $$
3
Hint. Observe that $$ \frac{\log 27}{\log m}=\frac{\log 3^3}{\log m}=3\cdot\log_m3=3\cdot b $$