How do you approximate log_7 36 given log_7 2=0.3562 and log_7 3=0.5646?

Oct 9, 2016

Let's first examine the prime factorization of $36$.

$36 = 6 \times 6 = 2 \times 3 \times 2 \times 3 = {2}^{2} \times {3}^{2}$

So, accordingly, we can write ${\log}_{7} \left(36\right)$ as ${\log}_{7} \left(2 \times 2 \times 3 \times 3\right)$, which by the sum rule of logarithms is equal to ${\log}_{7} \left(2\right) + {\log}_{7} \left(2\right) + {\log}_{7} \left(3\right) + {\log}_{7} \left(3\right)$.

This is equal to $0.3562 + 0.3562 + 0.5646 + 0.5646 = 1.8416$

Checking this answer using your calculator, you will find that our answer is quite close to the actual answer, ${\log}_{7} \left(36\right)$ having an approximate value of $1.8415644232$.

Hopefully this helps!