The answer to the present question is: log(a²b³/c⁴) = 2log(a) + 3log(b) - 4log(c)
For solving this kind of question, we are going to use these three identities:
log(x*y) = log(x) + log(y) ....(1)
log(x/y) = log(x) - log(y) ....(2)
log(xⁿ) = nlog(x) ....(3)
So, using (2), log(a²b³/c⁴) may be written as: log(a²b³) - log(c⁴) ....(4)
Again, using (1), log(a²b³) are often written as: log(a²) + log(b³) ....(5)
Now, Putting equation (4) in equation (3), we get:
log(a²) + log(b³) - log(c⁴) ....(5)
Now, using (3), expanding all the terms, we get:
log(a²) = 2log(a)log(b³) = 3log(b)log(c⁴) = 4log(c)
Putting of these in equation 5, we get:
2log(a) + 3log(b) - 4log(c)
Therefore, log(a²b³/c⁴) = 2log(a) +3log(b) - 4log(c)
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