From the rules of doing math using logarithms, we know the following
[tex]\begin{gathered} \text{log A + log B = log (AB)} \\ \text{log A - log B = log (A/B)} \\ \log A^B=B\cdot\log (A) \end{gathered}[/tex]
Then, using the data of your question we have the next equation
[tex]\begin{gathered} \log _511+\log _59=\log _5X \\ X=11\cdot9 \\ X=99 \\ \log _511+\log _59=\log _599 \end{gathered}[/tex]
The next one will give us the following:
[tex]\begin{gathered} \log _9Y-\log _97=\log _9\frac{8}{7} \\ \frac{8}{7}=\frac{Y}{7} \\ Y=8 \\ \log _98-\log _97=\log _9\frac{8}{7} \end{gathered}[/tex]
The last one will be as follows:
[tex]\begin{gathered} \log _2\frac{1}{25}=W\cdot\log _25 \\ 5^W=\frac{1}{25} \\ W=-2 \\ 5^{-2}=\frac{1}{5^2}=\frac{1}{25} \\ \log _2\frac{1}{25}=-2\cdot\log _25 \end{gathered}[/tex]
