Writing an equation as a single logorithm!! Help?

Question

Ok this equation is -2log3 1?x + 1/3log3 sqrt x


For some reason I keep getting 1/2log3x but I have a strong feeling that this is not correct at all can someone help me get on the right track please?

Answer 1

Is that log base 3? And is that 1/x?

-2 log[3] (1/x) + 1/3 log[3] (sqrt x)

Rewrite this as:
-2 log[3] (x^-1) + 1/3 log[3] (x^½)

Using the rule:
log x^k = k log x

-2 * -1 log[3] (x) + 1/3 * 1/2 log[3] (x)

Multiply:
2 log[3] (x) + (1/6) log[3] (x)

Factor a common log[3] (x)
(2 + 1/6) log[3] (x)

So there's your final answer, I believe:
2 1/6 log[3] x

Answer 2

-2log3 1?x + 1/3log3 sqrt x


what is "?" above?

Answer 3

-2(log3)1/x + 1/3(log3)x ^(1/2)
=lg(x^2) + lg(x^1/6) where lg = log3
=lg(x^(2 + 1/6) = lg(x^13/6)
=13/6 log3 (x)