Respuesta :

We have the following properties of logarithms:

[tex]\begin{gathered} \log (a\cdot b)=\log (a)+\log (b) \\ \log (\frac{a}{b})=\log (a)-\log (b) \\ \log (a^b)=b\cdot\log (a) \end{gathered}[/tex]

in this case we have the following expression:

[tex]\log _3y+7\log _3m-5\log _3y[/tex]

using the third property, we get:

[tex]\log _3y+7\log _3m-5\log _3y=\log _3y+\log _3m^7-\log _3y^5[/tex]

next, we can use the first property on the first two summands to get the following:

[tex]\log _3y+\log _3m^7-\log y^5=\log _3(ym^7)-\log _3y^5[/tex]

finally, using the second property, we get:

[tex]\log _3(ym^7)-\log _3y^5=\log _3(\frac{ym^7}{y^5})=\log _3(\frac{m^7}{y^4})[/tex]

therefore, the expresson as a single logarithm is:

[tex]\log _3(\frac{m^7}{y^4})[/tex]

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