It is given that $\log_8 = 1.893$, correct to 3 decimal places - HSC - SSCE Mathematics Extension 1 - Question 2 - 2017 - Paper 1

We know that:

$\log_4 = \frac{\log_8}{\log_4(8)}$

Firstly, we express $\log_4(8)$ as:

$\log_4(8) = \log_4(2^3) = 3 \cdot \log_4(2)$

Using the change of base formula:

$\log_4(2) = \frac{\log_2(2)}{\log_2(4)} = \frac{1}{2}$

Then:

$\log_4(8) = 3 \cdot \frac{1}{2} = \frac{3}{2} = 1.5$

Now we can find $\log_4$:

$\log_4 = \frac{\log_8}{\log_4(8)} = \frac{1.893}{1.5} = 1.262$

Rounded to two decimal places, this gives us:

$\log_4 \approx 1.26$