We can show these statements by numerical instances:
Lets solve a numerical example of the power property.
[tex]\log _a(u)^n=n\cdot\log _a(u)[/tex]where a is the base of the logarithm.
If we choose a=3, u=9 and n=2, we have on the left hand side:
[tex]\log _3(9)^2=\log _381=4[/tex]while on the right hand side, we obtain
[tex]2\cdot\log _39=2\cdot2=4[/tex]By comparing both results, we can see that the left hand side is equal to the right hand side, so the power propertuy is correct.
Lets take now the Quotient property:
[tex]\log _a(u)-\log _a(v)=\log _a(\frac{u}{v})[/tex]If we choose a=3, u =5 and v= 2, on the left hand side we have
[tex]\begin{gathered} \log _3(5)-\log _3(2)=1.4649-0.6309 \\ \log _3(5)-\log _3(2)=0.834 \end{gathered}[/tex]and on the right hand side of the property, we have
[tex]\log _3(\frac{5}{2})=\log _32.5=0.834[/tex]and we can see that both side coincide, then the property is correct.
1. Explain why log_a(1) =0.
We know that
[tex]x^0=1[/tex]for any x.
If we apply logarithm on both sides, we have
[tex]\log _ax^0=\log _a1[/tex]If we apply the power property on the left hand side, we have
[tex]0\cdot\log _ax[/tex]which is equal to zero. This means that
[tex]0=\log _a1[/tex]2. Explain why log_a a^x=x is true.
By means of the power property, we have
[tex]\log _aa^x=x\cdot\log _aa[/tex]but
[tex]\log _aa\text{ =1}[/tex]because the base (a) is the same as the number into the logarithm. Then, by substituting this result into the above equality, we have
[tex]\log _aa^x=x\cdot\log _aa=x\cdot1=x[/tex]then
[tex]\log _aa^x=x[/tex]