21. 7 \cdot \log_4 u - 3 \cdot \log_4 v^2

Key concept: Use the power rule and the quotient rule for logarithms to combine the expression into a single log.

Step by step

Start with the power rule:

$7\log_4 u=\log_4(u^7)$

and

$3\log_4 v^2=\log_4\big((v^2)^3\big)=\log_4(v^6)$

So the expression becomes

$\log_4(u^7)-\log_4(v^6)$

Now use the quotient rule:

$\log_4\left(\frac{u^7}{v^6}\right)$

So the simplified result is

$\boxed{\log_4\left(\frac{u^7}{v^6}\right)}$

assuming $u>0$ and $v\ne 0$.

Pitfall alert

Do not write $3\log_4 v^2=\log_4 v^6$ without thinking about parentheses. The correct step is $3\log_4(v^2)=\log_4((v^2)^3)=\log_4(v^6)$. The exponent applies to the whole $v^2$ term.

Try different conditions

If the expression were $7\log_4 u+3\log_4 v^2$, then the result would be

$\log_4(u^7v^6)$

If it were $7\log_4 u-\log_4 v^2$, then it would simplify to

$\log_4\left(\frac{u^7}{v^2}\right)$

Further reading

logarithm rules, power rule, quotient rule