Math / Algebra

QuestionWrite the expression as a single logarithm with coefficient 1 . Assume all variables represent positive real numbers. 5) $\frac{2}{9} \log _{n} 4 y+\frac{8}{7} \log _{n}\left(16 y^{2}\right)$ 5) $\qquad$ A) $\log _{n}\left(64 y^{9 / 10}\right)$ B) $\log _{n}\left(4^{39 / 20} y^{63 / 40}\right)$ C) $\log _{n}\left(4^{158 / 63} y^{158 / 63}\right)$ D) $\log _{n}\left(64 y^{39 / 40}\right)$

Studdy Solution

STEP 1

What is this asking? We need to combine these two separate logarithms with fractional coefficients into a single logarithm with a coefficient of 1. Watch out! Don't forget those logarithm rules!
Also, fractional exponents can be tricky, so let's be extra careful with those calculations.

STEP 2

1. Use the power rule
2. Use the product rule
3. Simplify the expression

STEP 3

Remember the power rule: $n \cdot \log_b(a) = \log_b(a^n)$ .
Let's apply this to our first term.
We have $\frac{2}{9} \log_n(4y)$ , which becomes $\log_n((4y)^{\frac{2}{9}})$ .
See? We moved that fraction up as an exponent!

STEP 4

Now, let's do the same thing to the second term.
We have $\frac{8}{7} \log_n(16y^2)$ , which becomes $\log_n((16y^2)^{\frac{8}{7}})$ .
Awesome!

STEP 5

Another important rule: $\log_b(a) + \log_b(c) = \log_b(a \cdot c)$ .
So, we can combine our two terms from the previous step: $\log_n((4y)^{\frac{2}{9}}) + \log_n((16y^2)^{\frac{8}{7}})$ becomes $\log_n((4y)^{\frac{2}{9}} \cdot (16y^2)^{\frac{8}{7}})$ .
We're getting closer to a single logarithm!

STEP 6

Let's simplify that expression inside the logarithm.
We have $(4y)^{\frac{2}{9}} \cdot (16y^2)^{\frac{8}{7}}$ .
Remember that $16$ is $4^2$ , so we can rewrite this as $(4y)^{\frac{2}{9}} \cdot (4^2 y^2)^{\frac{8}{7}}$ .
Now, using the power of a product rule $(ab)^n = a^n b^n$ , we get $4^{\frac{2}{9}}y^{\frac{2}{9}} \cdot 4^{\frac{16}{7}} y^{\frac{16}{7}}$ .

STEP 7

Now, group the terms with the same base together: $4^{\frac{2}{9}} \cdot 4^{\frac{16}{7}} \cdot y^{\frac{2}{9}} \cdot y^{\frac{16}{7}}$ .
Using the rule $a^m \cdot a^n = a^{m+n}$ , we get $4^{\frac{2}{9} + \frac{16}{7}} \cdot y^{\frac{2}{9} + \frac{16}{7}}$ .

STEP 8

Let's calculate those exponents! $\frac{2}{9} + \frac{16}{7} = \frac{2 \cdot 7}{63} + \frac{16 \cdot 9}{63} = \frac{14}{63} + \frac{144}{63} = \frac{158}{63}$ .
So, our expression becomes $4^{\frac{158}{63}}y^{\frac{158}{63}}$ .

STEP 9

Putting it all together, we have $\log_n(4^{\frac{158}{63}}y^{\frac{158}{63}})$ .
And that's it!

STEP 10

Our final answer is C) $\log_n(4^{\frac{158}{63}}y^{\frac{158}{63}})$ .

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Studdy Solution

STEP 1

What is this asking? We need to combine these two separate logarithms with fractional coefficients into a single logarithm with a coefficient of 1. Watch out! Don't forget those logarithm rules!
Also, fractional exponents can be tricky, so let's be extra careful with those calculations.

STEP 2

1. Use the power rule
2. Use the product rule
3. Simplify the expression

STEP 3

Remember the power rule: $n \cdot \log_b(a) = \log_b(a^n)$ .
Let's apply this to our first term.
We have $\frac{2}{9} \log_n(4y)$ , which becomes $\log_n((4y)^{\frac{2}{9}})$ .
See? We moved that fraction up as an exponent!

STEP 4

Now, let's do the same thing to the second term.
We have $\frac{8}{7} \log_n(16y^2)$ , which becomes $\log_n((16y^2)^{\frac{8}{7}})$ .
Awesome!

STEP 5

Another important rule: $\log_b(a) + \log_b(c) = \log_b(a \cdot c)$ .
So, we can combine our two terms from the previous step: $\log_n((4y)^{\frac{2}{9}}) + \log_n((16y^2)^{\frac{8}{7}})$ becomes $\log_n((4y)^{\frac{2}{9}} \cdot (16y^2)^{\frac{8}{7}})$ .
We're getting closer to a single logarithm!

STEP 6

STEP 7

Now, group the terms with the same base together: $4^{\frac{2}{9}} \cdot 4^{\frac{16}{7}} \cdot y^{\frac{2}{9}} \cdot y^{\frac{16}{7}}$ .
Using the rule $a^m \cdot a^n = a^{m+n}$ , we get $4^{\frac{2}{9} + \frac{16}{7}} \cdot y^{\frac{2}{9} + \frac{16}{7}}$ .

STEP 8

Let's calculate those exponents! $\frac{2}{9} + \frac{16}{7} = \frac{2 \cdot 7}{63} + \frac{16 \cdot 9}{63} = \frac{14}{63} + \frac{144}{63} = \frac{158}{63}$ .
So, our expression becomes $4^{\frac{158}{63}}y^{\frac{158}{63}}$ .

STEP 9

Putting it all together, we have $\log_n(4^{\frac{158}{63}}y^{\frac{158}{63}})$ .
And that's it!

STEP 10

Our final answer is C) $\log_n(4^{\frac{158}{63}}y^{\frac{158}{63}})$ .

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Studdy Solution

STEP 1

What is this asking? We need to combine these two separate logarithms with fractional coefficients into a *single* logarithm with a coefficient of **1**. Watch out! Don't forget those logarithm rules!Also, fractional exponents can be tricky, so let's be extra careful with those calculations.

STEP 2

1. Use the power rule2. Use the product rule3. Simplify the expression

STEP 3

STEP 4

Now, let's do the same thing to the second term.We have 87log⁡n(16y2)\frac{8}{7} \log_n(16y^2)78​logn​(16y2), which becomes log⁡n((16y2)87)\log_n((16y^2)^{\frac{8}{7}})logn​((16y2)78​).Awesome!

STEP 5

STEP 6

STEP 7

STEP 8

STEP 9

Putting it all together, we have log⁡n(415863y15863)\log_n(4^{\frac{158}{63}}y^{\frac{158}{63}})logn​(463158​y63158​).And that's it!

STEP 10

Our final answer is C) log⁡n(415863y15863)\log_n(4^{\frac{158}{63}}y^{\frac{158}{63}})logn​(463158​y63158​).

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We need to combine these two separate logarithms with fractional coefficients into a single logarithm with a coefficient of 1. Watch out! Don't forget those logarithm rules!
Also, fractional exponents can be tricky, so let's be extra careful with those calculations.

1. Use the power rule
2. Use the product rule
3. Simplify the expression

Now, let's do the same thing to the second term.
We have $\frac{8}{7} \log_n(16y^2)$ , which becomes $\log_n((16y^2)^{\frac{8}{7}})$ .
Awesome!

Putting it all together, we have $\log_n(4^{\frac{158}{63}}y^{\frac{158}{63}})$ .
And that's it!

Our final answer is C) $\log_n(4^{\frac{158}{63}}y^{\frac{158}{63}})$ .