Question Solve for $\gamma$ and $y$ in the equation $4^{6 \gamma} = 8^{2 y + 4}$ .

Studdy Solution

STEP 1

Assumptions
1. We are given the equation $4^{6\gamma} = 8^{2y+4}$ .
2. We need to solve for $\gamma$ in terms of $y$ .
3. We recognize that both 4 and 8 are powers of 2, which will allow us to simplify the equation using properties of exponents.

STEP 2

First, we express both sides of the equation in terms of a common base. In this case, we can express 4 and 8 as powers of 2.
$4 = 2^2 \quad \text{and} \quad 8 = 2^3$

STEP 3

Now, we rewrite the given equation using these expressions for 4 and 8.
$(2^2)^{6\gamma} = (2^3)^{2y+4}$

STEP 4

Next, we apply the power of a power rule for exponents, which states that $(a^m)^n = a^{mn}$ .
$2^{2 \cdot 6\gamma} = 2^{3 \cdot (2y+4)}$

STEP 5

We simplify the exponents by performing the multiplication.
$2^{12\gamma} = 2^{6y+12}$

STEP 6

Since the bases are the same and the equation is an equality, we can set the exponents equal to each other. This gives us the following equation:
$12\gamma = 6y + 12$

STEP 7

To solve for $\gamma$ , we need to isolate $\gamma$ on one side of the equation. We can do this by dividing both sides of the equation by 12.
$\gamma = \frac{6y + 12}{12}$

STEP 8

Simplify the right side of the equation by dividing each term by 12.
$\gamma = \frac{6y}{12} + \frac{12}{12}$

STEP 9

Reduce the fractions to their simplest form.
$\gamma = \frac{1}{2}y + 1$
So, the solution to the equation $4^{6\gamma} = 8^{2y+4}$ is
$\gamma = \frac{1}{2}y + 1$

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Question Solve for $\gamma$ and $y$ in the equation $4^{6 \gamma} = 8^{2 y + 4}$ .

Studdy Solution

STEP 1

Assumptions
1. We are given the equation $4^{6\gamma} = 8^{2y+4}$ .
2. We need to solve for $\gamma$ in terms of $y$ .
3. We recognize that both 4 and 8 are powers of 2, which will allow us to simplify the equation using properties of exponents.

STEP 2

First, we express both sides of the equation in terms of a common base. In this case, we can express 4 and 8 as powers of 2.
$4 = 2^2 \quad \text{and} \quad 8 = 2^3$

STEP 3

Now, we rewrite the given equation using these expressions for 4 and 8.
$(2^2)^{6\gamma} = (2^3)^{2y+4}$

STEP 4

Next, we apply the power of a power rule for exponents, which states that $(a^m)^n = a^{mn}$ .
$2^{2 \cdot 6\gamma} = 2^{3 \cdot (2y+4)}$

STEP 5

We simplify the exponents by performing the multiplication.
$2^{12\gamma} = 2^{6y+12}$

STEP 6

Since the bases are the same and the equation is an equality, we can set the exponents equal to each other. This gives us the following equation:
$12\gamma = 6y + 12$

STEP 7

To solve for $\gamma$ , we need to isolate $\gamma$ on one side of the equation. We can do this by dividing both sides of the equation by 12.
$\gamma = \frac{6y + 12}{12}$

STEP 8

Simplify the right side of the equation by dividing each term by 12.
$\gamma = \frac{6y}{12} + \frac{12}{12}$

STEP 9

Reduce the fractions to their simplest form.
$\gamma = \frac{1}{2}y + 1$
So, the solution to the equation $4^{6\gamma} = 8^{2y+4}$ is
$\gamma = \frac{1}{2}y + 1$

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Question Solve for γ\gammaγ and yyy in the equation 46γ=82y+44^{6 \gamma} = 8^{2 y + 4}46γ=82y+4.

Studdy Solution

STEP 1

Assumptions1. We are given the equation 46γ=82y+44^{6\gamma} = 8^{2y+4}46γ=82y+4.2. We need to solve for γ\gammaγ in terms of yyy.3. We recognize that both 4 and 8 are powers of 2, which will allow us to simplify the equation using properties of exponents.

STEP 2

First, we express both sides of the equation in terms of a common base. In this case, we can express 4 and 8 as powers of 2.4=22and8=234 = 2^2 \quad \text{and} \quad 8 = 2^34=22and8=23

STEP 3

Now, we rewrite the given equation using these expressions for 4 and 8.(22)6γ=(23)2y+4 (2^2)^{6\gamma} = (2^3)^{2y+4} (22)6γ=(23)2y+4

STEP 4

Next, we apply the power of a power rule for exponents, which states that (am)n=amn(a^m)^n = a^{mn}(am)n=amn.22⋅6γ=23⋅(2y+4) 2^{2 \cdot 6\gamma} = 2^{3 \cdot (2y+4)} 22⋅6γ=23⋅(2y+4)

STEP 5

We simplify the exponents by performing the multiplication.212γ=26y+12 2^{12\gamma} = 2^{6y+12} 212γ=26y+12

STEP 6

Since the bases are the same and the equation is an equality, we can set the exponents equal to each other. This gives us the following equation:12γ=6y+12 12\gamma = 6y + 12 12γ=6y+12

STEP 7

To solve for γ\gammaγ, we need to isolate γ\gammaγ on one side of the equation. We can do this by dividing both sides of the equation by 12.γ=6y+1212 \gamma = \frac{6y + 12}{12} γ=126y+12​

STEP 8

Simplify the right side of the equation by dividing each term by 12.γ=6y12+1212 \gamma = \frac{6y}{12} + \frac{12}{12} γ=126y​+1212​

STEP 9

Reduce the fractions to their simplest form.γ=12y+1 \gamma = \frac{1}{2}y + 1 γ=21​y+1So, the solution to the equation 46γ=82y+44^{6\gamma} = 8^{2y+4}46γ=82y+4 isγ=12y+1 \gamma = \frac{1}{2}y + 1 γ=21​y+1

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Question Solve for $\gamma$ and $y$ in the equation $4^{6 \gamma} = 8^{2 y + 4}$ .

Assumptions
1. We are given the equation $4^{6\gamma} = 8^{2y+4}$ .
2. We need to solve for $\gamma$ in terms of $y$ .
3. We recognize that both 4 and 8 are powers of 2, which will allow us to simplify the equation using properties of exponents.

First, we express both sides of the equation in terms of a common base. In this case, we can express 4 and 8 as powers of 2.
$4 = 2^2 \quad \text{and} \quad 8 = 2^3$

Now, we rewrite the given equation using these expressions for 4 and 8.
$(2^2)^{6\gamma} = (2^3)^{2y+4}$

Next, we apply the power of a power rule for exponents, which states that $(a^m)^n = a^{mn}$ .
$2^{2 \cdot 6\gamma} = 2^{3 \cdot (2y+4)}$

We simplify the exponents by performing the multiplication.
$2^{12\gamma} = 2^{6y+12}$

Since the bases are the same and the equation is an equality, we can set the exponents equal to each other. This gives us the following equation:
$12\gamma = 6y + 12$

To solve for $\gamma$ , we need to isolate $\gamma$ on one side of the equation. We can do this by dividing both sides of the equation by 12.
$\gamma = \frac{6y + 12}{12}$

Simplify the right side of the equation by dividing each term by 12.
$\gamma = \frac{6y}{12} + \frac{12}{12}$

Reduce the fractions to their simplest form.
$\gamma = \frac{1}{2}y + 1$
So, the solution to the equation $4^{6\gamma} = 8^{2y+4}$ is
$\gamma = \frac{1}{2}y + 1$