log problem

[Math Tutor]

If [tex]log_{a}b=-3[/tex]
find the value of the logarithm:
[tex]log_{b}(a^{9}b^{6})[/tex]

[mathbuddy]

As [tex]log_{a}b=-3[/tex]

Therefore [tex]log_{b}a=-1/3[/tex] [Using flipping the base formula]

Now [tex]log_{b}(a^{9}b^{6})[/tex]

= [tex]9log_{b}a + 6log_{b}b[/tex]

= 9(-1/3) + 6 [Because [tex]log_{b}a=-1/3[/tex]]

= -3 + 6
= 3

[Math Tutor]

Simplify the logarithmic expression:
[tex]M = \left(\log_b^4{a}+\log_a^4{b}+2\right)^{\frac{1}{2}}-\log_b{a}-\log_a{b}[/tex]

[burgess]

Yes this answer is absolutely correct!!!

As [tex]log_{a}b=-3[/tex]

Therefore [tex]log_{b}a=-1/3[/tex] [Using flipping the base formula]

Now [tex]log_{b}(a^{9}b^{6})[/tex]

= [tex]9log_{b}a + 6log_{b}b[/tex]

= 9(-1/3) + 6 [Because [tex]log_{b}a=-1/3[/tex]]

= -3 + 6
= 3

[HallsofIvy]

If [tex]log_{a}b=-3[/tex]

So [tex]b= a^{-3}= \frac{1}{a^3}[/tex] and then [tex]a= \frac{1}{b^{1/3}}[/tex]

find the value of the logarithm:
[tex]log_{b}(a^{9}b^{6})[/tex]

[tex]a^9= (1/b^{1/3})^9= \frac{1}{b^3}[/tex]

[tex]a^9b^6= \frac{b^6}{b^3}= b^6[/tex]
[tex]log_b(a^9b^6)= log_b(b^6)= 6[/tex].

[Guest]

I hate it when I make these silly typos!
Obviously [tex]\frac{b^6}{b^3}= b^3[/tex] so that [tex]log_b(a^9b^6)= 3[/tex], not 6.