Question $\log _{2} \sqrt[3]{2} \cdot \log _{2} \frac{1}{2}$

Studdy Solution

STEP 1

What is this asking? We're asked to multiply two logarithms, both with base 2, but with different arguments: the cube root of 2, and the reciprocal of 2. Watch out! Don't forget those logarithm rules!
Remembering how fractional exponents relate to roots is key here.

STEP 2

1. Simplify the First Logarithm
2. Simplify the Second Logarithm
3. Multiply the Simplified Logarithms

STEP 3

Alright, let's break down this first logarithm, $\log_{2} \sqrt[3]{2}$ .
Remember that the cube root of a number is the same as raising that number to the power of $\frac{1}{3}$ .
So, $\sqrt[3]{2}$ is the same as $2^{\frac{1}{3}}$ .
This makes our first logarithm $\log_{2} 2^{\frac{1}{3}}$ .

STEP 4

Now, we can use the super handy logarithm power rule!
This rule says $\log_{b} a^c = c \cdot \log_{b} a$ .
So, we can bring that $\frac{1}{3}$ exponent down in front, giving us $\frac{1}{3} \cdot \log_{2} 2$ .

STEP 5

And since $\log_{b} b = 1$ (because $b^1 = b$ ), we know that $\log_{2} 2 = 1$ .
So, our first logarithm simplifies to $\frac{1}{3} \cdot 1 = \frac{1}{3}$ .
Awesome!

STEP 6

Time for the second logarithm, $\log_{2} \frac{1}{2}$ .
Remember that $\frac{1}{x}$ is the same as $x^{-1}$ , so we can rewrite this as $\log_{2} 2^{-1}$ .

STEP 7

Let's use that power rule again!
We can bring the $-1$ exponent down in front, giving us $-1 \cdot \log_{2} 2$ .

STEP 8

And just like before, we know $\log_{2} 2 = 1$ .
So, our second logarithm simplifies to $-1 \cdot 1 = -1$ .
Fantastic!

STEP 9

We've simplified our original expression to $\frac{1}{3} \cdot -1$ .
Now, all we have to do is multiply these two numbers together.

STEP 10

Multiplying $\frac{1}{3}$ by $-1$ gives us $-\frac{1}{3}$ .
And that's our final answer!

STEP 11

$-\frac{1}{3}$

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