Question

(i) $\frac{2^{x^2}}{2^{2x}}=\frac{8^x}{64}$

Answer 1

x=2, 3

Answer 2

To solve the equation $\frac{2^{x^2}}{2^{2x}}=\frac{8^x}{64}$, we will use the properties of exponents to simplify both sides of the equation to the same base.

Step 1: Express all terms with the same base (base 2). We know that $8 = 2^3$ and $64 = 2^6$.

The left side of the equation is: $\frac{2^{x^2}}{2^{2x}}$ Using the exponent property $\frac{a^m}{a^n} = a^{m-n}$, we get: $2^{x^2 - 2x}$

The right side of the equation is: $\frac{8^x}{64}$ Substitute $8 = 2^3$ and $64 = 2^6$: $\frac{(2^3)^x}{2^6}$ Using the exponent property $(a^m)^n = a^{mn}$, we get: $\frac{2^{3x}}{2^6}$ Now, using the exponent property $\frac{a^m}{a^n} = a^{m-n}$, we get: $2^{3x - 6}$

Step 2: Equate the simplified expressions. Now the equation becomes: $2^{x^2 - 2x} = 2^{3x - 6}$

Step 3: Equate the exponents. Since the bases are the same (both are 2), the exponents must be equal: $x^2 - 2x = 3x - 6$

Step 4: Rearrange into a standard quadratic equation. Move all terms to one side to form a quadratic equation $ax^2 + bx + c = 0$: $x^2 - 2x - 3x + 6 = 0$ $x^2 - 5x + 6 = 0$

Step 5: Solve the quadratic equation. We can solve this quadratic equation by factoring. We need two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3. So, the equation can be factored as: $(x - 2)(x - 3) = 0$

This gives two possible solutions for $x$: $x - 2 = 0 \implies x = 2$ $x - 3 = 0 \implies x = 3$

Step 6: Verify the solutions (optional but recommended). For $x=2$: LHS: $\frac{2^{2^2}}{2^{2 \times 2}} = \frac{2^4}{2^4} = 1$ RHS: $\frac{8^2}{64} = \frac{64}{64} = 1$ LHS = RHS, so $x=2$ is a solution.

For $x=3$: LHS: $\frac{2^{3^2}}{2^{2 \times 3}} = \frac{2^9}{2^6} = 2^{9-6} = 2^3 = 8$ RHS: $\frac{8^3}{64} = \frac{512}{64} = 8$ LHS = RHS, so $x=3$ is a solution.

Both solutions are valid.

The final answer is $\boxed{x=2, 3}$.

Explanation of the solution: The given exponential equation is simplified by expressing all terms with the same base (base 2). Using exponent properties $a^m/a^n = a^{m-n}$ and $(a^m)^n = a^{mn}$, the equation transforms into $2^{x^2 - 2x} = 2^{3x - 6}$. Equating the exponents yields a quadratic equation $x^2 - 2x = 3x - 6$, which simplifies to $x^2 - 5x + 6 = 0$. Factoring this quadratic equation gives $(x-2)(x-3)=0$, leading to the solutions $x=2$ and $x=3$.