<s>[INST] Solving logarithm leaving in terms of $p$ and $q$ I would like to check the steps if Part a) is done correctly.
For Part b), how do I continue from below? I seem to stuck for $\log_{10}(5)$…
Here is the problem:

Given that $p = \log_{10} 2$ and $q = \log_{10} 7$, express the following in terms of $p$ and $q$.
a) $\log_{7} 4 = \frac{\log_{10} 4}{\log_{10} 7} = \frac{2 \log_{10} 2}{\log_{10} 7} = \frac{2p}{q}$
b) $\log_{10} \sqrt[3]{\frac{25}{49}} = \log_{10}5^\frac{2}{3} - \log_{10}7^\frac{2}{3} = \frac{2}{3}\log_{10}5 - \frac{2q}{3}$

[Source]
 [/INST] Part a) is perfect (nice job)
For part b) the $\log_{10} 5$ is throwing a monkey wrench and you can't express $5$ in terms of $2$ and $7$ using only multiplication and division and exponents.
BUt you have two other values at your disposal:  $\log_{10} 10 = 1$ and $\log_{10} 1 = 0$.
Can you express $5$ in terms of $2, 7, 10, 1$ in terms of multiplication and division and exponents.  (Hint: $1$ is fairly useless as it is the multiplicative identity)
Hint:

 $5 = \frac {10}2$

Solution:

 $\log_{10} 5 = \log_{10} \frac {10}2 = \log_{10} 10 - \log_{10} 2 = 1 - p$ so $\log_{10} \sqrt[3]{\frac {25}{49}}=\frac 23\log_{10} 5-\frac {2q}3 = \frac 23(1-p)-\frac {2q}3$

Alternatively:

 $\log_{10} \sqrt[3]{\frac {25}{49}}=\frac 23\log \frac 57=\frac 23\log \frac {10}{2*7} = \frac 23(\log 10 - (\log 2 + \log 7)) = \frac 23(1-p-q)$

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