QuestionSimplify. Enter the result as a single logarithm with a coefficient of 1. To enter the log, click in the answer cell, then click on the small yellow arrow on the right. Then click the Functions tab to access the log. If needed, you can change the base. Using the keyboard, you can also enter $\log _{3}\left(\frac{x^{2}}{2}\right)$ as $\log _{3} 3\left(x^{\wedge} 2 / 2\right)$ . $\log _{9}\left(6 x^{2}\right)-\log _{9}(7 x)=$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to combine two logarithms with the same base into a single logarithm, also with a coefficient of 1. Watch out! Remember the logarithm rules: $\log_b(x) - \log_b(y) = \log_b(\frac{x}{y})$ .
Also, don't forget to simplify the fraction inside the logarithm as much as possible!

STEP 2

1. Combine the logarithms.
2. Simplify the fraction.

STEP 3

We have $\log _{9}\left(6 x^{2}\right)-\log _{9}(7 x)$ .
Since the bases are the same, we can use the logarithm subtraction rule: $\log_b(x) - \log_b(y) = \log_b(\frac{x}{y})$ .
So, we get: $\log _{9}\left(6 x^{2}\right)-\log _{9}(7 x) = \log_9\left(\frac{6x^2}{7x}\right)$ This is awesome because we've combined the two logarithms into one!

STEP 4

Inside the logarithm, we have the fraction $\frac{6x^2}{7x}$ .
We can simplify this by dividing both the numerator and the denominator by $x$ .
Remember that $x^2 = x \cdot x$ , so dividing $x^2$ by $x$ gives us $x$ .
Dividing $x$ by itself gives us $1$ .
So, we get: $\frac{6x^2}{7x} = \frac{6x \cdot x}{7x} = \frac{6x}{7}$

STEP 5

Now, we can put this simplified fraction back into our logarithm: $\log_9\left(\frac{6x^2}{7x}\right) = \log_9\left(\frac{6x}{7}\right)$ Look at that, a single logarithm with a coefficient of 1!

STEP 6

Our final answer is $\log_9\left(\frac{6x}{7}\right)$ .

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