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To solve the equation [tex]\(8^{y+1} - 2^{3y+1} = 48\)[/tex], let's follow a structured, step-by-step approach. Here is the detailed solution:
### Step 1: Express Both Terms in the Same Base
Since [tex]\(8\)[/tex] can be expressed as a power of [tex]\(2\)[/tex] (specifically, [tex]\(8 = 2^3\)[/tex]), let's rewrite the equation using base 2.
[tex]\[ 8^{y+1} = (2^3)^{y+1} = 2^{3(y+1)} = 2^{3y+3} \][/tex]
So the original equation becomes:
[tex]\[ 2^{3y+3} - 2^{3y+1} = 48 \][/tex]
### Step 2: Simplify the Expression
We can factor out [tex]\(2^{3y+1}\)[/tex] from both terms on the left-hand side of the equation:
[tex]\[ 2^{3y+1}(2^2 - 1) = 48 \][/tex]
[tex]\[ 2^{3y+1} \times (4 - 1) = 48 \][/tex]
[tex]\[ 2^{3y+1} \times 3 = 48 \][/tex]
### Step 3: Isolate the Exponential Term
Now, isolate [tex]\(2^{3y+1}\)[/tex] by dividing both sides of the equation by 3:
[tex]\[ 2^{3y+1} = \frac{48}{3} \][/tex]
[tex]\[ 2^{3y+1} = 16 \][/tex]
### Step 4: Solve for the Exponent
Since [tex]\(16\)[/tex] is a power of 2 (specifically, [tex]\(16 = 2^4\)[/tex]), we can set the exponents equal to each other, because the bases are the same:
[tex]\[ 3y+1 = 4 \][/tex]
### Step 5: Solve for [tex]\(y\)[/tex]
Solve the equation [tex]\(3y + 1 = 4\)[/tex] for [tex]\(y\)[/tex]:
[tex]\[ 3y + 1 = 4 \][/tex]
[tex]\[ 3y = 4 - 1 \][/tex]
[tex]\[ 3y = 3 \][/tex]
[tex]\[ y = 1 \][/tex]
Thus, the real solution to the equation [tex]\(8^{y+1} - 2^{3y+1} = 48\)[/tex] is:
[tex]\[ y = 1 \][/tex]
### Step 6: Consider Complex Solutions
The general solutions often include complex values. Given that we can also represent logarithmic forms involving imaginary parts (but not required here), typically we might obtain results like:
[tex]\[ y \in \left\{1, \frac{\log(8) - 2i\pi}{3\log(2)}, \frac{\log(8) + 2i\pi}{3\log(2)} \right\} \][/tex]
Which simplifies to:
[tex]\[ y \in \left\{1, \frac{3\log(2) - 2i\pi}{3\log(2)}, \frac{3\log(2) + 2i\pi}{3\log(2)} \right\} \][/tex]
Upon simplification, we indeed get:
[tex]\[ y = 1 \][/tex]
and pseudo-periodic complex solutions in terms of [tex]\( \log(2) \)[/tex].
### Conclusion
Therefore, the solutions to the equation [tex]\(8^{y+1} - 2^{3y+1} = 48\)[/tex] are the following:
[tex]\[ y = 1 \][/tex]
and the additional complex solutions which stem from periodic properties of logarithmic and exponential functions:
[tex]\[ y = \frac{\log(8) - 2i\pi}{3\log(2)}, \quad y = \frac{\log(8) + 2i\pi}{3\log(2)} \][/tex]
### Step 1: Express Both Terms in the Same Base
Since [tex]\(8\)[/tex] can be expressed as a power of [tex]\(2\)[/tex] (specifically, [tex]\(8 = 2^3\)[/tex]), let's rewrite the equation using base 2.
[tex]\[ 8^{y+1} = (2^3)^{y+1} = 2^{3(y+1)} = 2^{3y+3} \][/tex]
So the original equation becomes:
[tex]\[ 2^{3y+3} - 2^{3y+1} = 48 \][/tex]
### Step 2: Simplify the Expression
We can factor out [tex]\(2^{3y+1}\)[/tex] from both terms on the left-hand side of the equation:
[tex]\[ 2^{3y+1}(2^2 - 1) = 48 \][/tex]
[tex]\[ 2^{3y+1} \times (4 - 1) = 48 \][/tex]
[tex]\[ 2^{3y+1} \times 3 = 48 \][/tex]
### Step 3: Isolate the Exponential Term
Now, isolate [tex]\(2^{3y+1}\)[/tex] by dividing both sides of the equation by 3:
[tex]\[ 2^{3y+1} = \frac{48}{3} \][/tex]
[tex]\[ 2^{3y+1} = 16 \][/tex]
### Step 4: Solve for the Exponent
Since [tex]\(16\)[/tex] is a power of 2 (specifically, [tex]\(16 = 2^4\)[/tex]), we can set the exponents equal to each other, because the bases are the same:
[tex]\[ 3y+1 = 4 \][/tex]
### Step 5: Solve for [tex]\(y\)[/tex]
Solve the equation [tex]\(3y + 1 = 4\)[/tex] for [tex]\(y\)[/tex]:
[tex]\[ 3y + 1 = 4 \][/tex]
[tex]\[ 3y = 4 - 1 \][/tex]
[tex]\[ 3y = 3 \][/tex]
[tex]\[ y = 1 \][/tex]
Thus, the real solution to the equation [tex]\(8^{y+1} - 2^{3y+1} = 48\)[/tex] is:
[tex]\[ y = 1 \][/tex]
### Step 6: Consider Complex Solutions
The general solutions often include complex values. Given that we can also represent logarithmic forms involving imaginary parts (but not required here), typically we might obtain results like:
[tex]\[ y \in \left\{1, \frac{\log(8) - 2i\pi}{3\log(2)}, \frac{\log(8) + 2i\pi}{3\log(2)} \right\} \][/tex]
Which simplifies to:
[tex]\[ y \in \left\{1, \frac{3\log(2) - 2i\pi}{3\log(2)}, \frac{3\log(2) + 2i\pi}{3\log(2)} \right\} \][/tex]
Upon simplification, we indeed get:
[tex]\[ y = 1 \][/tex]
and pseudo-periodic complex solutions in terms of [tex]\( \log(2) \)[/tex].
### Conclusion
Therefore, the solutions to the equation [tex]\(8^{y+1} - 2^{3y+1} = 48\)[/tex] are the following:
[tex]\[ y = 1 \][/tex]
and the additional complex solutions which stem from periodic properties of logarithmic and exponential functions:
[tex]\[ y = \frac{\log(8) - 2i\pi}{3\log(2)}, \quad y = \frac{\log(8) + 2i\pi}{3\log(2)} \][/tex]
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