Logarithm question?

Question

-2logbase3 1/x + 1/3 log base 3 sqrt x

I got log base 3 x^2 + log base 3 x ^ (1/6). Am I right??

Answer 1

So far so good. But you can do more:

logbase3 X^2 * X^1/6 = logbase3 x^13/6

Bozo

Answer 2

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

First, change the inside of the logs to having a power. Note that 1/x is equal to x^(-1) and sqrt(x) is equal to x^(1/2)

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

Now, we can move the powers outside of the logs, as per the log property log[base b](a^c) = c*log[base b](a).

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

We can then simplify the constants/numbers,

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

And we can add these logs normally. The adding isn't immediately obvious, but it will be when I factor the common factor of log[base 3](x) out.

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

Combining the constants, we have

log[base 3](x) [13/6], therefore, we have

(13/6) log[base 3](x)

as our final answer.

Answer 3

All logs are base 3. We have:

(-2)log(1/x) + (1/3)log(x^(1/2))
= (-2)(-log x) + (1/2)(1/3) log x
= 2 log x + (1/6) log x
= log x² + log x^(1/6)

So you are correct.