Rules of logarithms!! College Algebra!!!?

Question

Use the product, quotient, and power rules of logarithms to rewrite the expression as a single logarithm. Assume that all variables represent positive real numbers.
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1/2 log2 (x^4 )+ 1/4 log2(x^4) - 1/6 log2 x

Answer 1

Let log be log base 2 in the following:-
(1/2) log (x^4) + (1/4) log(x^4) - (1/6)log x
(3/4) log (x^4) - (1/6) log x
log x³ - (1/6) log x
3 log x - (1 / 6) log x
(17/6) log x

Answer 2

first the power rule...
The coeficcient in front of each log becomes a power of the argument...

log2 (x^4)^(1/2) + log2 (x^4)^(1/4) - log2 x^(1/6)

ok, now use the product rule to combine the logs being added by multiplying the arguments...

log2 x^2 + log2 x = log2 x^3
thus, we have log2 x^3 - log2 x^(1/6).

Now the quotient rule..
log2 (x^3) - log2 x^(1/6) = log2 (x^3)/ (x^(1/6))

All done : )

Answer 3

the rule of thumb is that (assuming they have a similar base) log(a) + log(b) = log(ab), and log(a) - log(b) = log(a/b). log(base 2)6 - log(base 2)15 + log(base 2)20 = log(base 2) (6*20/15) = log(base 2) (8) = 3

Answer 4

1/2 log(x^4) + 1/4(x^4) - 1/6(log x) (base 2 is not included)

log(x^4)^1/2 + log (x^4)^1/4 - log(x)^1/6 ( since a log b = log (b)^a)

log x^2 + log x - log(x)^1/6

log[(x^2)(x)/x^1/6)] (since log a + log b - log c = log(ab/c))

log(x^3/x^1/6)
log (x^(3 - 1/6)
log(x)^(17/6)