Question

Use an over/under table to estimate the solution to the exponential equation 3+2^3x=9 to two decimal places.

Answer 1

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

\[
2^{3x} = 9 - 3
\]
\[
2^{3x} = 6
\]

Next, we can take the logarithm of both sides. Let's use the base 2 logarithm for convenience:

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

Now we can solve for \(x\):

\[
x = \frac{\log_2(6)}{3}
\]

To estimate this value, we can use the change of base formula to convert the logarithm to a more familiar base, such as base 10:

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

Using known values:
- \(\log_{10}(6) \approx 0.7781\)
- \(\log_{10}(2) \approx 0.3010\)

Now we calculate \(\log_2(6)\):

\[
\log_2(6) \approx \frac{0.7781}{0.3010} \approx 2.585
\]

Now use this to find \(x\):

\[
x = \frac{\log_2(6)}{3} \approx \frac{2.585}{3} \approx 0.86167
\]

Now we round \(x\) to two decimal places:

\[
x \approx 0.86
\]

### Over/Under Table

To build an over/under table to refine our estimate, we will evaluate \(3 + 2^{3x}\) for values around \(x = 0.86\).

- If \(x = 0.8\):
\[
3 + 2^{3(0.8)} = 3 + 2^{2.4} \approx 3 + 5.278 \approx 8.278 \quad (\text{Under 9})
\]

- If \(x = 0.9\):
\[
3 + 2^{3(0.9)} = 3 + 2^{2.7} \approx 3 + 6.3496 \approx 9.3496 \quad (\text{Over 9})
\]

### Summary of Findings

- At \(x = 0.8\), \(3 + 2^{3x} \approx 8.278\) (Under 9)
- At \(x = 0.9\), \(3 + 2^{3x} \approx 9.3496\) (Over 9)

Thus, the value of \(x\) is between \(0.8\) and \(0.9\). To further narrow it down:

- If \(x = 0.85\):
\[
3 + 2^{3(0.85)} = 3 + 2^{2.55} \approx 3 + 5.656 \approx 8.656 \quad (\text{Under 9})
\]

Now we refine it to:

- If \(x = 0.88\):
\[
3 + 2^{3(0.88)} = 3 + 2^{2.64} \approx 3 + 6.196 \approx 9.196 \quad (\text{Over 9})
\]

### Final Range

From our evaluations:

- \(x = 0.85\) gives an under value, and \(x = 0.88\) gives an over value.

Hence, our solution narrows down to:

\[
\boxed{0.86}
\]

This confirms the value of \(x\) to two decimal places as \(0.86\).