Question

16^(2x-3)=(√8)^(x+1)

Answer 1

Ответ:

Здраствуйте, ответ указен на фотографии

Answer 2

To solve the equation $16^{2x-3} = (\sqrt{8})^{x+1}$, you can simplify both sides and then solve for $x$.

First, simplify both sides of the equation:

On the left side: $16^{2x-3} = (4^2)^{2x-3} = 4^{2(2x-3)} = 4^{4x-6}$

On the right side: $(\sqrt{8})^{x+1} = (2\sqrt{2})^{x+1} = 2^{x+1}(\sqrt{2})^{x+1} = 2^{x+1} \cdot 2^{\frac{1}{2}(x+1)} = 2^{x+1+\frac{1}{2}(x+1)} = 2^{x+\frac{3}{2}(x+1)}$

Now, the equation becomes:

$4^{4x-6} = 2^{x+\frac{3}{2}(x+1)}$

Since both sides have the same base (2), you can set the exponents equal to each other:

$4x-6 = x + \frac{3}{2}(x+1)$

Now, solve for $x$:

First, distribute $\frac{3}{2}$ on the right side:

$4x-6 = x + \frac{3}{2}x + \frac{3}{2}$

Combine like terms:

$4x - 6 = \frac{5}{2}x + \frac{3}{2}$

Now, isolate $x$:

Subtract $\frac{5}{2}x$ from both sides:

$4x - \frac{5}{2}x - 6 = \frac{3}{2}$

Combine like terms on the left side:

$\frac{3}{2}x - 6 = \frac{3}{2}$

Add 6 to both sides:

$\frac{3}{2}x = \frac{3}{2} + 6$

$\frac{3}{2}x = \frac{15}{2}$

Now, divide both sides by $\frac{3}{2}$ to solve for $x$:

$x = \frac{\frac{15}{2}}{\frac{3}{2}}$

Simplify:

$x = \frac{15}{3}$

$x = 5$

So, the solution to the equation $16^{2x-3} = (\sqrt{8})^{x+1}$ is $x = 5$.