How do you approximate log_7 16 given log_7 2=0.3562 and log_7 3=0.5646?

Nov 14, 2016

We have that $16 = {2}^{4}$, or $2 \times 2 \times 2 \times 2$. Writing in logarithms, we have:

${\log}_{7} \left(16\right) = {\log}_{7} \left(2 \times 2 \times 2 \times 2\right)$

Using the sum rule of logarithms that ${\log}_{a} n + {\log}_{a} m = {\log}_{a} \left(n \times m\right)$, we have:

$\implies {\log}_{7} 2 + {\log}_{7} 2 + {\log}_{7} 2 + {\log}_{7} 2$

$\implies 0.3562 + 0.3562 + 0.3562 + 0.3562$

$\implies 1.4248$

Hence, ${\log}_{7} 16 \cong 1.4248$. Checking with a calculator gives ${\log}_{7} 16 \cong 1.424828748$. So, our estimate was correct to 4 decimals.

Hopefully this helps!