Use the Laws of Logarithms to expand the expression:
\[\log_{3} (\frac{ x(x^2+10) }{ \sqrt{x^2-10} })\]
We can begin by applying the difference of logarithms with the same base: \[\log_{a} b - \log_{a} c \iff \log_{a} (\frac{ b }{ c })\] Applying the property to the exercise: \[\log_{3} (\frac{ x(x^2+10) }{ \sqrt{x^2-10} }) \iff \log_{3} (x(x^2+10))-\log_{a} (\sqrt{x^2-10})\]
ok I understand that. I'm confused where to go from there
The very strategy of expanding logarithms is reduced to adjusting the expressions in order to make them fit any law of logarithms you currently know. Let's take for example, getting rid of that square root. Of course, square roots are the contrary operation to exponents, but by this same logic, we can then utilize the exponential law of rational exponents in order to make it fir the law of logarithms we currently know, I'll write it here: \(\log_{a} b^n \iff (n).\log_{a} b\). in combination with: \(\sqrt[n]{a^m} \iff a ^{\frac{ m }{ n }}\) will allow us to get rid of the radical on the second logarithm: \[\log_{3} (x(x^2+10))-\log_{3} (\sqrt{x^2-10} \iff \log_{3} (x(x^2+10))-\log_{3} (x^2-10)^{\frac{ 1 }{ 2 }}\] To further obtain the expression: \[\log_{3} (x(x^2+10))-(\frac{ 1 }{ 2 })\log_{3} (x^2-10)\]
This is an example of what the thought process behind the expansion of a logarithm implies, which is expressing any term on it's own more simple logarithm.
alright so then you have to work inside the parentheses, right?
That's correct. And you continue doing that until you can't use the laws any further.
im confused with the first set of parentheses, but the second set would just be \[\frac{ x^2 }{ 10}\]
Show me your work.
\[\log_{3} (x(x^2+10)) = \log_{3} (x^3+10x) = \]
Yes, though I would suggest using the law of addition: \[\log_{a} b + \log_{a} c \iff \log_{a} b.c\]
so.. \[\log_{3}( x )+ \log_{3} (x^2+10)\]
correct.
so then the final answer is.. \[\log_{3} (x) + \log_{3} (x^2+10) - (\frac{ 1 }{ 2 }) \log_{3} (x^2-10)\]
That's correct.
thank you
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