Question

Use an over/under table to estimate the solution to the exponential equation 3+2^3x=9
to two decimal places.(1 point)
The solution to the exponential equation is x≈Remaining Attempts : 1

Answer 1

To solve the equation \( 3 + 2^{3x} = 9 \), we first isolate the exponential term:

\[
2^{3x} = 9 - 3
\]

\[
2^{3x} = 6
\]

Next, we take the logarithm of both sides to solve for \( x \):

\[
3x = \log_2(6)
\]

To compute \( \log_2(6) \), we can convert it using the change of base formula:

\[
\log_2(6) = \frac{\log_{10}(6)}{\log_{10}(2)}
\]

Calculating \( \log_{10}(6) \) and \( \log_{10}(2) \):

\[
\log_{10}(6) \approx 0.7781
\]
\[
\log_{10}(2) \approx 0.3010
\]

Now we can find \( \log_2(6) \):

\[
\log_2(6) \approx \frac{0.7781}{0.3010} \approx 2.58
\]

Now we can find \( x \):

\[
3x \approx 2.58 \implies x \approx \frac{2.58}{3} \approx 0.86
\]

### Over/Under Table Method

Let's check some values around \( x = 0.86 \) to narrow it down to two decimal places.

1. **For \( x = 0.80 \)**:

\[
3 + 2^{3(0.80)} = 3 + 2^{2.4} \approx 3 + 5.278 \approx 8.278 \quad (\text{too low})
\]

2. **For \( x = 0.90 \)**:

\[
3 + 2^{3(0.90)} = 3 + 2^{2.7} \approx 3 + 6.349 \approx 9.349 \quad (\text{too high})
\]

So, \( x = 0.80 \) is too low and \( x = 0.90 \) is too high.

3. **Now let's try \( x = 0.85 \)**:

\[
3 + 2^{3(0.85)} = 3 + 2^{2.55} \approx 3 + 5.656 \approx 8.656 \quad (\text{too low})
\]

4. **Then \( x = 0.88 \)**:

\[
3 + 2^{3(0.88)} = 3 + 2^{2.64} \approx 3 + 6.343 \approx 9.343 \quad (\text{too high})
\]

5. **Now \( x = 0.87 \)**:

\[
3 + 2^{3(0.87)} = 3 + 2^{2.61} \approx 3 + 5.744 \approx 8.744 \quad (\text{too low})
\]

Finally, we can now narrow down:

- \( x = 0.87 \) (too low)
- \( x = 0.88 \) (too high)

Thus, the solution to the exponential equation \( 3 + 2^{3x} = 9 \) is approximately:

\[
\boxed{0.87}
\]