Question

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



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**: \(x = \frac{19}{5}\).