求满足方程 $\left(\frac{1}{8}y^{3}\right)^{-1}\div\left(4y^{2}\right)^{\tfrac{1}{2}}=2^{a}y^{b}.$ 的整数 $a$ 和 $b$。 求满足方程 $\left(\frac{1}{8}y^{3}\right)^{-1}\div\left(4y^{2}\right)^{\tfrac{1}{2}}=2^{a}y^{b}.$ 的整数 $a$ 和 $b$。 解方程 $\log_{3}(x-8)=2-\dfrac{1}{\log_{x}3}$。 解方程 $\log_{3}(x-8)=2-\dfrac{1}{\log_{x}3}$。 解方程 $4^{x+1}+7\left(2^{x}\right)=2.$

1. Simplify the first term using negative index laws

$\left(\frac{1}{8}y^3\right)^{-1} = \left(2^{-3}y^3\right)^{-1} = 2^3 y^{-3} = 8y^{-3}$

2. Simplify the second term using fractional index laws

$\left(4y^2\right)^{\frac{1}{2}} = \left(2^2 y^2\right)^{\frac{1}{2}} = 2^1 y^1 = 2y$

3. Perform the division and equate to find a and b

$\left(\frac{1}{8}y^3\right)^{-1} \div \left(4y^2\right)^{\frac{1}{2}} = \frac{8y^{-3}}{2y} = 4y^{-4} = 2^2 y^{-4}$
Comparing with $2^a y^b$, we have $a = 2$ and $b = -4$.

1. Apply change of base to the term $\frac{1}{\log_{x}3}$

$\frac{1}{\log_{x}3} = \log_{3}x$

The equation becomes:
$\log_{3}(x-8) = 2 - \log_{3}x$

2. Group logarithmic terms and apply the product law of logarithms

$\log_{3}(x-8) + \log_{3}x = 2$
$\log_{3}(x(x-8)) = 2$

3. Convert the logarithmic equation to an exponential equation

$x(x-8) = 3^2$
$x^2 - 8x = 9$

4. Solve the resulting quadratic equation

$x^2 - 8x - 9 = 0$
$(x - 9)(x + 1) = 0$
$x = 9$ or $x = -1$

5. Check for valid solutions in the original logarithmic equation

For $x = -1$, $\log_{3}(-1-8)$ is undefined (argument must be positive).
Also, the base of $\log_{x}3$ must be positive and not equal to 1.
Therefore, $x = -1$ is rejected.
$x = 9$ is valid.

1. Express the equation in terms of $2^x$

$4(4^x) + 7(2^x) - 2 = 0$
$4(2^x)^2 + 7(2^x) - 2 = 0$

2. Use substitution to form and solve a quadratic equation

Let $u = 2^x$.
$4u^2 + 7u - 2 = 0$
$(4u - 1)(u + 2) = 0$
$u = \frac{1}{4} \text{ or } u = -2$

3. Solve for $x$ and reject invalid solutions

When $u = \frac{1}{4}$:
$2^x = \frac{1}{4}$
$2^x = 2^{-2}$
$x = -2$

When $u = -2$:
$2^x = -2$
(No real solution since $2^x > 0$ for all real $x$)