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 \(0.86\).