9log27X – 4log9X
the 27 and 9 that are between log and X are supposed to be subscripted
And please simplify it
a) log3X
b) log9X
C) log27X
d) 3/4 * log3X
e) 2/3 *log3X
Thanks. I don't no if there's a short cut to this since one base is 27 and the other is 9.
the 27 and 9 that are between log and X are supposed to be subscripted
And please simplify it
a) log3X
b) log9X
C) log27X
d) 3/4 * log3X
e) 2/3 *log3X
Thanks. I don't no if there's a short cut to this since one base is 27 and the other is 9.
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27 is 3³, and 9 is 3².
9 log(base 3³) x - 4 log(base 3²) x
log(base 3³) x^9 - log(base 3²) x^4
1/(log(base x^9) 3³) - 1/(log(base x^4) 3²)
1/(log(base x^9) 3³) - 1/(log(base x^4) 3²)
1/3 ∙ 1/(log(base x^9) 3) - 1/2 ∙ 1/(log(base x^4) 3)
(log(base 3) x^9)/3 - (log(base 3) x^4)/2
(2 log(base 3) x^9 - 3 log(base 3) x^4)/6
(log(base 3) x^18 - log(base 3) x^12)/6
(log(base 3) x^6)/6
6 (log(base 3) x)/6
(log(base 3) x)
Answer: A
9 log(base 3³) x - 4 log(base 3²) x
log(base 3³) x^9 - log(base 3²) x^4
1/(log(base x^9) 3³) - 1/(log(base x^4) 3²)
1/(log(base x^9) 3³) - 1/(log(base x^4) 3²)
1/3 ∙ 1/(log(base x^9) 3) - 1/2 ∙ 1/(log(base x^4) 3)
(log(base 3) x^9)/3 - (log(base 3) x^4)/2
(2 log(base 3) x^9 - 3 log(base 3) x^4)/6
(log(base 3) x^18 - log(base 3) x^12)/6
(log(base 3) x^6)/6
6 (log(base 3) x)/6
(log(base 3) x)
Answer: A
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I'm pretty sure there's an easier/less complicated way of doing it. I don't play around with logarithms for some months, so I just punched in all the properties I could remember, and manipulated from there. It is a valid and correct way of solving it, though.
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I'm glad I helped you, anyway!
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