Math / Algebra

QuestionWrite the expression as a single logarithm. $4 \log _{b}(9 y+1)+\frac{1}{2} \log _{b}(y+4)$ ${ }^{\log }(\square)$

Studdy Solution

STEP 1

What is this asking? Combine these two logarithms into one single logarithm! Watch out! Don't forget the logarithm power rule and the product rule!

STEP 2

1. Apply the power rule to the logarithms
2. Use the product rule to combine them

STEP 3

Alright, let's start by applying the power rule to each logarithm!
The power rule says that if you have a coefficient in front of a log, you can move it as an exponent inside the log.
So, for the first term $4 \log_b(9y+1)$ , we can rewrite it as:
$\log_b((9y+1)^4)$

STEP 4

Now, let's do the same for the second term $\frac{1}{2} \log_b(y+4)$ .
Using the power rule, we get:
$\log_b((y+4)^{1/2})$

STEP 5

Now that we've got our logs in a nice form, let's combine them using the product rule!
The product rule for logs says that when you add two logs with the same base, you can multiply their insides.
So, we have:
$\log_b((9y+1)^4) + \log_b((y+4)^{1/2}) = \log_b((9y+1)^4 \cdot (y+4)^{1/2})$

STEP 6

The expression as a single logarithm is:
$\log_b((9y+1)^4 \cdot (y+4)^{1/2})$

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QuestionWrite the expression as a single logarithm. $4 \log _{b}(9 y+1)+\frac{1}{2} \log _{b}(y+4)$ ${ }^{\log }(\square)$

Studdy Solution

STEP 1

What is this asking? Combine these two logarithms into one single logarithm! Watch out! Don't forget the logarithm power rule and the product rule!

STEP 2

1. Apply the power rule to the logarithms
2. Use the product rule to combine them

STEP 3

Alright, let's start by applying the power rule to each logarithm!
The power rule says that if you have a coefficient in front of a log, you can move it as an exponent inside the log.
So, for the first term $4 \log_b(9y+1)$ , we can rewrite it as:
$\log_b((9y+1)^4)$

STEP 4

Now, let's do the same for the second term $\frac{1}{2} \log_b(y+4)$ .
Using the power rule, we get:
$\log_b((y+4)^{1/2})$

STEP 5

Now that we've got our logs in a nice form, let's combine them using the product rule!
The product rule for logs says that when you add two logs with the same base, you can multiply their insides.
So, we have:
$\log_b((9y+1)^4) + \log_b((y+4)^{1/2}) = \log_b((9y+1)^4 \cdot (y+4)^{1/2})$

STEP 6

The expression as a single logarithm is:
$\log_b((9y+1)^4 \cdot (y+4)^{1/2})$

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QuestionWrite the expression as a single logarithm. 4log⁡b(9y+1)+12log⁡b(y+4)4 \log _{b}(9 y+1)+\frac{1}{2} \log _{b}(y+4)4logb​(9y+1)+21​logb​(y+4) log⁡(□){ }^{\log }(\square)log(□)

Studdy Solution

STEP 1

What is this asking? Combine these two logarithms into one single logarithm! Watch out! Don't forget the logarithm power rule and the product rule!

STEP 2

1. Apply the power rule to the logarithms2. Use the product rule to combine them

STEP 3

STEP 4

Now, let's do the same for the second term 12log⁡b(y+4)\frac{1}{2} \log_b(y+4)21​logb​(y+4).Using the power rule, we get:log⁡b((y+4)1/2)\log_b((y+4)^{1/2})logb​((y+4)1/2)

STEP 5

STEP 6

The expression as a single logarithm is:log⁡b((9y+1)4⋅(y+4)1/2)\log_b((9y+1)^4 \cdot (y+4)^{1/2})logb​((9y+1)4⋅(y+4)1/2)

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QuestionWrite the expression as a single logarithm. $4 \log _{b}(9 y+1)+\frac{1}{2} \log _{b}(y+4)$ ${ }^{\log }(\square)$

1. Apply the power rule to the logarithms
2. Use the product rule to combine them

Now, let's do the same for the second term $\frac{1}{2} \log_b(y+4)$ .
Using the power rule, we get:
$\log_b((y+4)^{1/2})$

The expression as a single logarithm is:
$\log_b((9y+1)^4 \cdot (y+4)^{1/2})$