Prove that; log_{8}(x)= 2/3 log_{4}(x) Hence find log_{8}(6) if log_{4}(3)=0.7925

To prove that $\log_{8}(x) = \frac{2}{3} \log_{4}(x)$, we start by expressing the logarithms in terms of a common base. Let's use base 2 for this purpose.

First, recall the change of base formula for logarithms: $\log_{a}(b) = \frac{\log_{c}(b)}{\log_{c}(a)}$

Using this formula, we can express $\log_{8}(x)$ and $\log_{4}(x)$ in terms of base 2:

$\log_{8}(x) = \frac{\log_{2}(x)}{\log_{2}(8)}$ $\log_{4}(x) = \frac{\log_{2}(x)}{\log_{2}(4)}$

Next, we need to find $\log_{2}(8)$ and $\log_{2}(4)$:

$\log_{2}(8) = \log_{2}(2^3) = 3$ $\log_{2}(4) = \log_{2}(2^2) = 2$

Substituting these values back into the expressions for $\log_{8}(x)$ and $\log_{4}(x)$, we get:

$\log_{8}(x) = \frac{\log_{2}(x)}{3}$ $\log_{4}(x) = \frac{\log_{2}(x)}{2}$

Now, we need to show that $\log_{8}(x) = \frac{2}{3} \log_{4}(x)$. Substitute the expressions we found:

$\log_{8}(x) = \frac{\log_{2}(x)}{3}$ $\frac{2}{3} \log_{4}(x) = \frac{2}{3} \left( \frac{\log_{2}(x)}{2} \right) = \frac{\log_{2}(x)}{3}$

Since both expressions are equal, we have:

$\log_{8}(x) = \frac{2}{3} \log_{4}(x)$

This completes the proof.

Next, we need to find $\log_{8}(6)$ given that $\log_{4}(3) = 0.7925$.

Using the proven relationship $\log_{8}(x) = \frac{2}{3} \log_{4}(x)$, we substitute $x = 6$:

$\log_{8}(6) = \frac{2}{3} \log_{4}(6)$

To find $\log_{4}(6)$, we use the property of logarithms that allows us to express $\log_{4}(6)$ in terms of known values:

$\log_{4}(6) = \log_{4}(3 \cdot 2) = \log_{4}(3) + \log_{4}(2)$

We know $\log_{4}(3) = 0.7925$. Now, we need to find $\log_{4}(2)$:

$\log_{4}(2) = \frac{\log_{2}(2)}{\log_{2}(4)} = \frac{1}{2}$

So,

$\log_{4}(6) = 0.7925 + 0.5 = 1.2925$

Now, substitute this back into the expression for $\log_{8}(6)$:

$\log_{8}(6) = \frac{2}{3} \times 1.2925 = \frac{2 \times 1.2925}{3} = \frac{2.585}{3} = 0.8617$

Therefore,

$\log_{8}(6) = 0.8617$