Solve the follow equation: 8^x+1= 16^2x-4

Question

A
x=5x=5x=5

B
x=1x=1x=1

C
x=1311x=\frac{13}{11}x=
11
13
​


D
x=195x=\frac{19}{5}x=
5
19
​

Answer 1

To solve the equation \(8^{x+1} = 16^{2x-4}\), we start by expressing both bases as powers of 2.

1. Rewrite \(8\) and \(16\): \[ 8 = 2^3 \quad \text{and} \quad 16 = 2^4 \] Thus, we can rewrite the equation as: \[ (2^3)^{x+1} = (2^4)^{2x-4} \]

2. Using the property of exponents \((a^m)^n = a^{m \cdot n}\): \[ 2^{3(x+1)} = 2^{4(2x-4)} \]

3. This simplifies to: \[ 2^{3x + 3} = 2^{8x - 16} \]

4. Since the bases are the same, we can set the exponents equal to each other: \[ 3x + 3 = 8x - 16 \]

5. Now, solve for \(x\): \[ 3 + 16 = 8x - 3x \] \[ 19 = 5x \] \[ x = \frac{19}{5} \]

Therefore, the solution is: \[ \boxed{\frac{19}{5}} \]

So the correct option is D: \(x = \frac{19}{5}\).