Thank you for the clarification. Let's correct the expression for Question 3 based on the updated information.

Revised Question 3:

Given Problem: Show that the equation $9^{x + \frac{1}{2}} - 7(3^{x+1}) - 55 = 0$ has only one solution and find its value correct to 2 significant figures.

Solution:

The given equation is: $9^{x + \frac{1}{2}} - 7(3^{x+1}) - 55 = 0$

Step 1: Simplify the equation

• Recall that $9 = 3^2$, so $9^{x + \frac{1}{2}} = (3^2)^{x + \frac{1}{2}} = 3^{2x + 1}$.

The equation now becomes: $3^{2x + 1} - 7(3^{x+1}) - 55 = 0$

Step 2: Substitute $y = 3^x$

Let's introduce a substitution $y = 3^x$. Then, $3^{2x} = y^2$, and the equation can be rewritten as: $3 \cdot y^2 - 7 \cdot 3 \cdot y - 55 = 0$

Simplify the equation: $3y^2 - 21y - 55 = 0$

Step 3: Solve the quadratic equation

This is a quadratic equation in terms of $y$. We can solve it using the quadratic formula: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 3$, $b = -21$, and $c = -55$.

Plugging in the values: $y = \frac{21 \pm \sqrt{(-21)^2 - 4(3)(-55)}}{2(3)}$ $y = \frac{21 \pm \sqrt{441 + 660}}{6}$ $y = \frac{21 \pm \sqrt{1101}}{6}$

Calculating the square root of 1101: $y = \frac{21 \pm 33.19}{6}$

This gives us two possible values for $y$:

1. $y = \frac{21 + 33.19}{6} \approx \frac{54.19}{6} \approx 9.03$
2. $y = \frac{21 - 33.19}{6} \approx \frac{-12.19}{6} \approx -2.03$ (which is not possible since $y = 3^x$ must be positive).

So, the only valid solution is $y \approx 9.03$.

Step 4: Solve for $x$

Recall that $y = 3^x$. Therefore: $3^x \approx 9.03$

Take the logarithm (base 3) of both sides: $x = \log_3(9.03)$

Convert this to the common logarithm (base 10): $x = \frac{\log_{10}(9.03)}{\log_{10}(3)} \approx \frac{0.955}{0.477} \approx 2.0$

Final Answer:

The solution is $x \approx 2.0$ correct to 2 significant figures.

If you have any questions about these steps or want further details, feel free to ask!

Follow-Up Questions:

1. Would you like a detailed explanation of the quadratic formula used here?
2. Would you like to know how to solve similar equations involving exponential functions?
3. Do you want a more in-depth explanation of converting logarithms between different bases?
4. How would you check if the quadratic equation has extraneous solutions?
5. Are you interested in exploring why the negative solution for $y$ was discarded?

Tip:

When solving exponential equations, always consider using logarithms to simplify the process and isolate the variable.