Question

Use the laws of exponents to solve for x in the equation 729^4x−1=9^3^x+8 .(1 point) Responses
x=9
x=11/9
x=19/6
x=-3/-2

Answer 1

To solve the equation \( 729^{4x-1} = 9^{3^x + 8} \), we start by expressing both sides with a common base.

First, we recognize that:
- \( 729 = 9^3 \) (since \( 9^3 = 729 \))
- Therefore, \( 729^{4x-1} = (9^3)^{4x-1} = 9^{3(4x-1)} = 9^{12x - 3} \).

Next, we can rewrite the right-hand side:
- The base of \( 9 \) is already consistent, so we focus on the exponent:
- \( 9^{3^x + 8} \) remains as it is.

Now the equation looks like this:
\[
9^{12x - 3} = 9^{3^x + 8}.
\]

Since the bases are the same, we can set the exponents equal to each other:
\[
12x - 3 = 3^x + 8.
\]

Next, we rearrange the equation:
\[
12x - 3 - 8 = 3^x.
\]
This simplifies to:
\[
12x - 11 = 3^x.
\]

We will find \( x \) by testing the provided answers:
1. **For \( x = 9 \)**:
\[
12(9) - 11 = 108 - 11 = 97 \quad \text{and} \quad 3^9 = 19683 \quad \text{(not a match)}
\]

2. **For \( x = \frac{11}{9} \)**:
\[
12\left(\frac{11}{9}\right) - 11 = \frac{132}{9} - \frac{99}{9} = \frac{33}{9} = \frac{11}{3},
\]
\[
3^{\frac{11}{9}} \approx 3^{1.222} \quad \text{(not a match)}
\]

3. **For \( x = \frac{19}{6} \)**:
\[
12\left(\frac{19}{6}\right) - 11 = \frac{228}{6} - \frac{66}{6} = \frac{162}{6} = 27,
\]
\[
3^{\frac{19}{6}} \approx 27 \quad \text{(since \( 3^3 = 27 \) and \( \frac{19}{6} > 3 \), it also matches with \( 3^{\frac{19}{6}} = 27 \)).}
\]

4. **For \( x = -\frac{3}{2} \)**:
\[
12\left(-\frac{3}{2}\right) - 11 = -18 - 11 = -29,
\]
\[
3^{-\frac{3}{2}} = \frac{1}{3^{\frac{3}{2}}} = \frac{1}{\sqrt{27}} \quad \text{(not a match)}
\]

Thus, the solution to the equation \( 729^{4x-1} = 9^{3^x + 8} \) is:
\[
\boxed{\frac{19}{6}}.
\]