Math / Algebra

QuestionÉtant donné $a=\log _{b}(16)$
Exprime $\log _{b}(8 b)$ en fonction de a

Studdy Solution

STEP 1

What is this asking? We're trying to rewrite $\log_b(8b)$ using $a$ , where $a = \log_b(16)$ . Watch out! Remember your logarithm rules!
Don't mix up bases, and be careful with how you rewrite the arguments.

STEP 2

1. Rewrite the given equation
2. Rewrite the target expression
3. Express the target in terms of $a$

STEP 3

We're given $a = \log_b(16)$ .
Since $16 = 2^4$ , we can rewrite this as $a = \log_b(2^4)$ .

STEP 4

Using the power rule of logarithms, which states that $\log_b(x^y) = y \cdot \log_b(x)$ , we can bring the exponent down: $a = 4 \cdot \log_b(2)$ .

STEP 5

To isolate $\log_b(2)$ , we can divide both sides by 4: $\log_b(2) = \frac{a}{4}$ .
This will be useful later!

STEP 6

We want to express $\log_b(8b)$ in terms of $a$ .
We can rewrite $8$ as $2^3$ , and we can use the fact that $8b$ is the same as $8 \cdot b$ .
This gives us $\log_b(2^3 \cdot b)$ .

STEP 7

The product rule of logarithms says $\log_b(x \cdot y) = \log_b(x) + \log_b(y)$ .
Applying this to our expression gives us $\log_b(2^3) + \log_b(b)$ .

STEP 8

We know that $\log_b(b) = 1$ for any base $b$ .
Also, using the power rule again (like we did earlier), we get $3 \cdot \log_b(2) + 1$ .

STEP 9

Remember how we found that $\log_b(2) = \frac{a}{4}$ ?
Let's substitute that into our expression: $3 \cdot \frac{a}{4} + 1$ .

STEP 10

This simplifies to $\frac{3a}{4} + 1$ .
We can also write it as $\frac{3a + 4}{4}$ if we want a single fraction.

STEP 11

We found that $\log_b(8b)$ can be expressed as $\frac{3a}{4} + 1$ or $\frac{3a + 4}{4}$ in terms of $a$ .

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QuestionÉtant donné $a=\log _{b}(16)$
Exprime $\log _{b}(8 b)$ en fonction de a

Studdy Solution

STEP 1

What is this asking? We're trying to rewrite $\log_b(8b)$ using $a$ , where $a = \log_b(16)$ . Watch out! Remember your logarithm rules!
Don't mix up bases, and be careful with how you rewrite the arguments.

STEP 2

1. Rewrite the given equation
2. Rewrite the target expression
3. Express the target in terms of $a$

STEP 3

We're given $a = \log_b(16)$ .
Since $16 = 2^4$ , we can rewrite this as $a = \log_b(2^4)$ .

STEP 4

Using the power rule of logarithms, which states that $\log_b(x^y) = y \cdot \log_b(x)$ , we can bring the exponent down: $a = 4 \cdot \log_b(2)$ .

STEP 5

To isolate $\log_b(2)$ , we can divide both sides by 4: $\log_b(2) = \frac{a}{4}$ .
This will be useful later!

STEP 6

We want to express $\log_b(8b)$ in terms of $a$ .
We can rewrite $8$ as $2^3$ , and we can use the fact that $8b$ is the same as $8 \cdot b$ .
This gives us $\log_b(2^3 \cdot b)$ .

STEP 7

The product rule of logarithms says $\log_b(x \cdot y) = \log_b(x) + \log_b(y)$ .
Applying this to our expression gives us $\log_b(2^3) + \log_b(b)$ .

STEP 8

We know that $\log_b(b) = 1$ for any base $b$ .
Also, using the power rule again (like we did earlier), we get $3 \cdot \log_b(2) + 1$ .

STEP 9

Remember how we found that $\log_b(2) = \frac{a}{4}$ ?
Let's substitute that into our expression: $3 \cdot \frac{a}{4} + 1$ .

STEP 10

This simplifies to $\frac{3a}{4} + 1$ .
We can also write it as $\frac{3a + 4}{4}$ if we want a single fraction.

STEP 11

We found that $\log_b(8b)$ can be expressed as $\frac{3a}{4} + 1$ or $\frac{3a + 4}{4}$ in terms of $a$ .

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QuestionÉtant donné a=log⁡b(16)a=\log _{b}(16)a=logb​(16)Exprime log⁡b(8b)\log _{b}(8 b)logb​(8b) en fonction de a

Studdy Solution

STEP 1

What is this asking? We're trying to rewrite log⁡b(8b)\log_b(8b)logb​(8b) using aaa, where a=log⁡b(16)a = \log_b(16)a=logb​(16). Watch out! Remember your logarithm rules!Don't mix up bases, and be careful with how you rewrite the arguments.

STEP 2

1. Rewrite the given equation2. Rewrite the target expression3. Express the target in terms of aaa

STEP 3

We're given a=log⁡b(16)a = \log_b(16)a=logb​(16).Since 16=2416 = 2^416=24, we can rewrite this as a=log⁡b(24)a = \log_b(2^4)a=logb​(24).

STEP 4

Using the power rule of logarithms, which states that log⁡b(xy)=y⋅log⁡b(x)\log_b(x^y) = y \cdot \log_b(x)logb​(xy)=y⋅logb​(x), we can bring the exponent down: a=4⋅log⁡b(2)a = 4 \cdot \log_b(2)a=4⋅logb​(2).

STEP 5

To isolate log⁡b(2)\log_b(2)logb​(2), we can divide both sides by **4**: log⁡b(2)=a4\log_b(2) = \frac{a}{4}logb​(2)=4a​.This will be useful later!

STEP 6

We want to express log⁡b(8b)\log_b(8b)logb​(8b) in terms of aaa.We can rewrite 888 as 232^323, and we can use the fact that 8b8b8b is the same as 8⋅b8 \cdot b8⋅b.This gives us log⁡b(23⋅b)\log_b(2^3 \cdot b)logb​(23⋅b).

STEP 7

The product rule of logarithms says log⁡b(x⋅y)=log⁡b(x)+log⁡b(y)\log_b(x \cdot y) = \log_b(x) + \log_b(y)logb​(x⋅y)=logb​(x)+logb​(y).Applying this to our expression gives us log⁡b(23)+log⁡b(b)\log_b(2^3) + \log_b(b)logb​(23)+logb​(b).

STEP 8

We know that log⁡b(b)=1\log_b(b) = 1logb​(b)=1 for any base bbb.Also, using the power rule again (like we did earlier), we get 3⋅log⁡b(2)+13 \cdot \log_b(2) + 13⋅logb​(2)+1.

STEP 9

Remember how we found that log⁡b(2)=a4\log_b(2) = \frac{a}{4}logb​(2)=4a​?Let's substitute that into our expression: 3⋅a4+13 \cdot \frac{a}{4} + 13⋅4a​+1.

STEP 10

This simplifies to 3a4+1\frac{3a}{4} + 143a​+1.We can also write it as 3a+44\frac{3a + 4}{4}43a+4​ if we want a single fraction.

STEP 11

We found that log⁡b(8b)\log_b(8b)logb​(8b) can be expressed as 3a4+1\frac{3a}{4} + 143a​+1 or 3a+44\frac{3a + 4}{4}43a+4​ in terms of aaa.

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QuestionÉtant donné $a=\log _{b}(16)$
Exprime $\log _{b}(8 b)$ en fonction de a

What is this asking? We're trying to rewrite $\log_b(8b)$ using $a$ , where $a = \log_b(16)$ . Watch out! Remember your logarithm rules!
Don't mix up bases, and be careful with how you rewrite the arguments.

1. Rewrite the given equation
2. Rewrite the target expression
3. Express the target in terms of $a$

We're given $a = \log_b(16)$ .
Since $16 = 2^4$ , we can rewrite this as $a = \log_b(2^4)$ .

Using the power rule of logarithms, which states that $\log_b(x^y) = y \cdot \log_b(x)$ , we can bring the exponent down: $a = 4 \cdot \log_b(2)$ .

To isolate $\log_b(2)$ , we can divide both sides by 4: $\log_b(2) = \frac{a}{4}$ .
This will be useful later!

We want to express $\log_b(8b)$ in terms of $a$ .
We can rewrite $8$ as $2^3$ , and we can use the fact that $8b$ is the same as $8 \cdot b$ .
This gives us $\log_b(2^3 \cdot b)$ .

The product rule of logarithms says $\log_b(x \cdot y) = \log_b(x) + \log_b(y)$ .
Applying this to our expression gives us $\log_b(2^3) + \log_b(b)$ .

We know that $\log_b(b) = 1$ for any base $b$ .
Also, using the power rule again (like we did earlier), we get $3 \cdot \log_b(2) + 1$ .

Remember how we found that $\log_b(2) = \frac{a}{4}$ ?
Let's substitute that into our expression: $3 \cdot \frac{a}{4} + 1$ .

This simplifies to $\frac{3a}{4} + 1$ .
We can also write it as $\frac{3a + 4}{4}$ if we want a single fraction.

We found that $\log_b(8b)$ can be expressed as $\frac{3a}{4} + 1$ or $\frac{3a + 4}{4}$ in terms of $a$ .