Math / Algebra

Question $\log _{9}\left(3 x^{2}\right)+\log _{9}\left(4 x^{6}\right)$

Studdy Solution

STEP 1

What is this asking? We're asked to simplify the sum of two logarithms with the same base. Watch out! Remember the logarithm product rule!
Also, don't forget to simplify as much as possible.

STEP 2

1. Combine the logarithms.
2. Simplify the expression.

STEP 3

Since the logarithms have the same base, we can use the product rule: $\log_b(m) + \log_b(n) = \log_b(m \cdot n)$ .
In our case, $b = 9$ , $m = 3x^2$ , and $n = 4x^6$ .
So, we have: $\log_9(3x^2) + \log_9(4x^6) = \log_9((3x^2) \cdot (4x^6))$

STEP 4

Let's multiply the terms inside the logarithm: $\log_9((3x^2) \cdot (4x^6)) = \log_9(3 \cdot 4 \cdot x^2 \cdot x^6) = \log_9(12x^8)$ Remember, when multiplying like bases, we add the exponents: $x^2 \cdot x^6 = x^{2+6} = x^8$ .

STEP 5

We can rewrite the argument of the logarithm, $12x^8$ , as $3 \cdot 4 \cdot x^8$ .
This gives us: $\log_9(12x^8) = \log_9(3 \cdot 4 \cdot x^8)$

STEP 6

We can apply the product rule in reverse to separate the logarithm: $\log_9(3 \cdot 4 \cdot x^8) = \log_9(3) + \log_9(4) + \log_9(x^8)$

STEP 7

Since $9 = 3^2$ , we have $\log_9(3) = \frac{1}{2}$ because $9^{\frac{1}{2}} = 3$ .

STEP 8

We can simplify $\log_9(x^8)$ using the power rule: $\log_b(m^n) = n \cdot \log_b(m)$ .
In our case, $b = 9$ , $m = x$ , and $n = 8$ .
So, we get: $\log_9(x^8) = 8\log_9(x)$

STEP 9

Putting it all together, we have: $\log_9(3) + \log_9(4) + \log_9(x^8) = \frac{1}{2} + \log_9(4) + 8\log_9(x)$

STEP 10

Our simplified expression is $\frac{1}{2} + \log_9(4) + 8\log_9(x)$ .

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Question $\log _{9}\left(3 x^{2}\right)+\log _{9}\left(4 x^{6}\right)$

Studdy Solution

STEP 1

What is this asking? We're asked to simplify the sum of two logarithms with the same base. Watch out! Remember the logarithm product rule!
Also, don't forget to simplify as much as possible.

STEP 2

1. Combine the logarithms.
2. Simplify the expression.

STEP 3

Since the logarithms have the same base, we can use the product rule: $\log_b(m) + \log_b(n) = \log_b(m \cdot n)$ .
In our case, $b = 9$ , $m = 3x^2$ , and $n = 4x^6$ .
So, we have: $\log_9(3x^2) + \log_9(4x^6) = \log_9((3x^2) \cdot (4x^6))$

STEP 4

Let's multiply the terms inside the logarithm: $\log_9((3x^2) \cdot (4x^6)) = \log_9(3 \cdot 4 \cdot x^2 \cdot x^6) = \log_9(12x^8)$ Remember, when multiplying like bases, we add the exponents: $x^2 \cdot x^6 = x^{2+6} = x^8$ .

STEP 5

We can rewrite the argument of the logarithm, $12x^8$ , as $3 \cdot 4 \cdot x^8$ .
This gives us: $\log_9(12x^8) = \log_9(3 \cdot 4 \cdot x^8)$

STEP 6

We can apply the product rule in reverse to separate the logarithm: $\log_9(3 \cdot 4 \cdot x^8) = \log_9(3) + \log_9(4) + \log_9(x^8)$

STEP 7

Since $9 = 3^2$ , we have $\log_9(3) = \frac{1}{2}$ because $9^{\frac{1}{2}} = 3$ .

STEP 8

We can simplify $\log_9(x^8)$ using the power rule: $\log_b(m^n) = n \cdot \log_b(m)$ .
In our case, $b = 9$ , $m = x$ , and $n = 8$ .
So, we get: $\log_9(x^8) = 8\log_9(x)$

STEP 9

Putting it all together, we have: $\log_9(3) + \log_9(4) + \log_9(x^8) = \frac{1}{2} + \log_9(4) + 8\log_9(x)$

STEP 10

Our simplified expression is $\frac{1}{2} + \log_9(4) + 8\log_9(x)$ .

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Questionlog⁡9(3x2)+log⁡9(4x6)\log _{9}\left(3 x^{2}\right)+\log _{9}\left(4 x^{6}\right)log9​(3x2)+log9​(4x6)

Studdy Solution

STEP 1

What is this asking? We're asked to simplify the sum of two logarithms with the same base. Watch out! Remember the logarithm product rule!Also, don't forget to simplify as much as possible.

STEP 2

1. Combine the logarithms.2. Simplify the expression.

STEP 3

STEP 4

STEP 5

We can rewrite the argument of the logarithm, 12x812x^812x8, as 3⋅4⋅x83 \cdot 4 \cdot x^83⋅4⋅x8.This gives us: log⁡9(12x8)=log⁡9(3⋅4⋅x8) \log_9(12x^8) = \log_9(3 \cdot 4 \cdot x^8) log9​(12x8)=log9​(3⋅4⋅x8)

STEP 6

We can apply the product rule in reverse to separate the logarithm: log⁡9(3⋅4⋅x8)=log⁡9(3)+log⁡9(4)+log⁡9(x8) \log_9(3 \cdot 4 \cdot x^8) = \log_9(3) + \log_9(4) + \log_9(x^8) log9​(3⋅4⋅x8)=log9​(3)+log9​(4)+log9​(x8)

STEP 7

Since 9=329 = 3^29=32, we have log⁡9(3)=12\log_9(3) = \frac{1}{2}log9​(3)=21​ because 912=39^{\frac{1}{2}} = 3921​=3.

STEP 8

We can simplify log⁡9(x8)\log_9(x^8)log9​(x8) using the power rule: log⁡b(mn)=n⋅log⁡b(m)\log_b(m^n) = n \cdot \log_b(m)logb​(mn)=n⋅logb​(m).In our case, b=9b = 9b=9, m=xm = xm=x, and n=8n = 8n=8.So, we get: log⁡9(x8)=8log⁡9(x) \log_9(x^8) = 8\log_9(x) log9​(x8)=8log9​(x)

STEP 9

Putting it all together, we have: log⁡9(3)+log⁡9(4)+log⁡9(x8)=12+log⁡9(4)+8log⁡9(x) \log_9(3) + \log_9(4) + \log_9(x^8) = \frac{1}{2} + \log_9(4) + 8\log_9(x) log9​(3)+log9​(4)+log9​(x8)=21​+log9​(4)+8log9​(x)

STEP 10

Our simplified expression is 12+log⁡9(4)+8log⁡9(x)\frac{1}{2} + \log_9(4) + 8\log_9(x)21​+log9​(4)+8log9​(x).

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Question $\log _{9}\left(3 x^{2}\right)+\log _{9}\left(4 x^{6}\right)$

What is this asking? We're asked to simplify the sum of two logarithms with the same base. Watch out! Remember the logarithm product rule!
Also, don't forget to simplify as much as possible.

1. Combine the logarithms.
2. Simplify the expression.

We can rewrite the argument of the logarithm, $12x^8$ , as $3 \cdot 4 \cdot x^8$ .
This gives us: $\log_9(12x^8) = \log_9(3 \cdot 4 \cdot x^8)$

We can apply the product rule in reverse to separate the logarithm: $\log_9(3 \cdot 4 \cdot x^8) = \log_9(3) + \log_9(4) + \log_9(x^8)$

Since $9 = 3^2$ , we have $\log_9(3) = \frac{1}{2}$ because $9^{\frac{1}{2}} = 3$ .

We can simplify $\log_9(x^8)$ using the power rule: $\log_b(m^n) = n \cdot \log_b(m)$ .
In our case, $b = 9$ , $m = x$ , and $n = 8$ .
So, we get: $\log_9(x^8) = 8\log_9(x)$

Putting it all together, we have: $\log_9(3) + \log_9(4) + \log_9(x^8) = \frac{1}{2} + \log_9(4) + 8\log_9(x)$

Our simplified expression is $\frac{1}{2} + \log_9(4) + 8\log_9(x)$ .