To solve the exponential equation \( 3 + 23^x = 9 \), let's first isolate \( 23^x \):
\[ 23^x = 9 - 3 \] \[ 23^x = 6 \]
Next, we can take the logarithm of both sides to solve for \( x \):
\[ x = \log_{23}(6) \]
Using the change of base formula:
\[ x = \frac{\log(6)}{\log(23)} \]
Now we can compute this using logarithm values. Calculating these using a calculator:
\[ \log(6) \approx 0.7781 \] \[ \log(23) \approx 1.3617 \]
Now to divide these:
\[ x \approx \frac{0.7781}{1.3617} \approx 0.571 \]
Now, we need to confirm this using an over/under table. We'll check values around \( x \approx 0.57 \):
Estimate Table
| \( x \) | \( 23^x \) | \( 3 + 23^x \) | Comparison with 9 | |---------|------------|----------------|-------------------| | 0.50 | \( 23^{0.50} \approx 4.796 \) | \( 3 + 4.796 \approx 7.796 \) | Under | | 0.60 | \( 23^{0.60} \approx 6.616 \) | \( 3 + 6.616 \approx 9.616 \) | Over |
From the table:
- For \( x = 0.50 \), \( 3 + 23^{0.50} \) gives approximately 7.80 (which is under 9).
- For \( x = 0.60 \), \( 3 + 23^{0.60} \) gives approximately 9.62 (which is over 9).
This indicates that the solution \( x \) is between 0.50 and 0.60. To narrow it down further, we can try an average of these two:
Narrowing Down Further
-
Try \( x = 0.55 \):
\[ 23^{0.55} \approx 5.696 \rightarrow 3 + 5.696 \approx 8.696 \quad (\text{Under}) \]
-
Try \( x = 0.57 \):
\[ 23^{0.57} \approx 6.049 \rightarrow 3 + 6.049 \approx 9.049 \quad (\text{Over}) \]
So we can conclude that:
\[ x \text{ is between } 0.55 \text{ and } 0.57. \]
-
Try \( x = 0.56 \):
\[ 23^{0.56} \approx 5.866 \rightarrow 3 + 5.866 \approx 8.866 \quad (\text{Under}) \]
Now we can establish:
- \( 0.56 \) is under 9.
- \( 0.57 \) is over 9.
Using interpolation, we can conclude that a more precise estimate is:
- The solution to the exponential equation is approximately \( x \approx 0.57 \).
Final Answer
Thus, the final answer is:
\[ x \approx 0.57. \]