Let's solve the two problems from the image.

Question 4:

We have the equation: $64^{-3x + 6} = \frac{16^{4x - 3}}{\sqrt{2^{2x + 36}}}$

Let's express all numbers with the base 2:

• $64 = 2^6$, so $64^{-3x + 6} = (2^6)^{-3x + 6} = 2^{6(-3x + 6)} = 2^{-18x + 36}$.
• $16 = 2^4$, so $16^{4x - 3} = (2^4)^{4x - 3} = 2^{16x - 12}$.
• $\sqrt{2^{2x + 36}} = 2^{(2x + 36)/2} = 2^{x + 18}$.

Substitute these back into the equation: $2^{-18x + 36} = \frac{2^{16x - 12}}{2^{x + 18}}$

Simplify the right-hand side: $2^{-18x + 36} = 2^{(16x - 12) - (x + 18)} = 2^{16x - 12 - x - 18} = 2^{15x - 30}$

Since the bases are the same, set the exponents equal to each other: $-18x + 36 = 15x - 30$

Solve for $x$: $36 + 30 = 15x + 18x \implies 66 = 33x \implies x = 2$

So, the answer is A. 2.

Question 5:

We have the equation: $\frac{3^x}{27^{x+2}} = 9 \sqrt{3^{x - 3} \cdot 3^{x + 5}}$

Convert all terms to base 3:

• $27 = 3^3$, so $27^{x+2} = (3^3)^{x+2} = 3^{3(x+2)} = 3^{3x+6}$.
• $9 = 3^2$, so $9 = 3^2$.

Rewrite the equation using these bases: $\frac{3^x}{3^{3x + 6}} = 3^2 \sqrt{3^{x - 3 + x + 5}}$

Simplify the left side: $3^{x - (3x + 6)} = 3^{x - 3x - 6} = 3^{-2x - 6}$

Simplify the right side: $3^2 \cdot \sqrt{3^{2x + 2}} = 3^2 \cdot 3^{(2x + 2)/2} = 3^2 \cdot 3^{x + 1} = 3^{2 + x + 1} = 3^{x + 3}$

Now we have: $3^{-2x - 6} = 3^{x + 3}$

Set the exponents equal to each other: $-2x - 6 = x + 3$

Solve for $x$: $-2x - x = 3 + 6 \implies -3x = 9 \implies x = -3$

So, the answer is A. -3.

Conclusion:

• Question 4: A. 2
• Question 5: A. -3

Would you like more details or have any other questions?

Related Questions:

1. How do you solve exponential equations with different bases?
2. What are some methods to simplify expressions with exponents?
3. How can logarithms be used to solve exponential equations?
4. What are the properties of exponents that help in solving such equations?
5. How does changing the base of an exponent affect the solution of an equation?

Tip:

When solving equations involving exponents, always look for a common base to simplify the equation.