log 3 36 - 4log 9 30 + 16log 81 (sqrt)15

Question

anonymous · Answer

$$\log_{3}36-4\log_{9}30+16\log_{81}\sqrt{15}   $$

australopithecus · Answer

you just want to combine these right?

australopithecus · Answer

or write as one term

australopithecus · Answer

ugh equation editor isnt working one sec

australopithecus · Answer

$$\log(a^b) = b \log(a)$$

australopithecus · Answer

$$-\log(a) = \log(a^{-1}) = \log(\frac{1}{a})$$

$$\log_c(a) + \log_c(b) = \log(ab)$$

Note however,
$$\log_c(a) + \log_d(b) \neq \log(ab)$$
as they are different bases

australopithecus · Answer

can you solve it now?

australopithecus · Answer

Note that one rule isnt unique to logarithms 
$$a^{-c} = \frac{1}{a^c}$$
$$\frac{1}{a^{-c}} = a^{c}$$

One more note,

$$a^1 = a$$

australopithecus · Answer

For the last term in your problem it may be a bit tricky first off note that,

$$\sqrt{15} = 15^{\frac{1}{2}}$$

australopithecus · Answer

so you will need to multiply 16 by 1/2

australopithecus · Answer

as a general rule I suppose to make things simple,

$$\log(a) - \log(b) = \log(\frac{a}{b})$$

anonymous · Answer

i dont think I have difficulty with the first two terms, but the last one I've been trying to use the change of base formula