Question

Use an over/under table to estimate the solution to the exponential equation 3+23x=9 to two decimal places.(1 point) The solution to the exponential equation is x≈

Answer 1

To solve the equation \(3 + 23^x = 9\), we first isolate the exponential term:

\[
23^x = 9 - 3
\]
\[
23^x = 6
\]

Next, we can take the logarithm of both sides:

\[
x \ln(23) = \ln(6)
\]

Now, we can solve for \(x\):

\[
x = \frac{\ln(6)}{\ln(23)}
\]

Now we can estimate \(x\) using the logarithm values approximately:

- \(\ln(6) \approx 1.791759\)
- \(\ln(23) \approx 3.044522\)

Now substitute these values back into the equation for \(x\):

\[
x \approx \frac{1.791759}{3.044522} \approx 0.587
\]

For better precision, we can also calculate \(x\) using a calculator:

Using a calculator for more accurate values:
\[
\ln(6) \approx 1.791759469228055
\]
\[
\ln(23) \approx 3.044522437723423
\]

Calculating \(x\):

\[
x \approx \frac{1.791759469228055}{3.044522437723423} \approx 0.587
\]

Now, to refine our estimate using an over/under table:

1. Try \(x = 0.5\):
\[
23^{0.5} = \sqrt{23} \approx 4.7958,
\]
\[
3 + 4.7958 \approx 7.7958 < 9.
\]

2. Try \(x = 0.6\):
\[
23^{0.6} \approx 5.569,
\]
\[
3 + 5.569 \approx 8.569 < 9.
\]

3. Try \(x = 0.7\):
\[
23^{0.7} \approx 6.272,
\]
\[
3 + 6.272 \approx 9.272 > 9.
\]

Now we have narrowed it down:
- For \(x = 0.6\), \(7.7958 < 9\).
- For \(x = 0.7\), \(9.272 > 9\).

Let's try a value in between, such as \(x = 0.65\):
\[
23^{0.65} \approx 5.897,
\]
\[
3 + 5.897 \approx 8.897 < 9.
\]

Now try \(x = 0.68\):
\[
23^{0.68} \approx 6.136,
\]
\[
3 + 6.136 \approx 9.136 > 9.
\]

So, we can conclude that the solution is between \(0.65\) and \(0.68\).

Finally, through additional iterations or using a more precise logarithmic value, we can find \(x\) on the calculator. Based on the over/under method, we can narrow it down further.

Thus, we estimate that:

\[
x \approx 0.64
\]

So the final answer to two decimal places is:

\[
\boxed{0.64}
\]