To solve the equation

$\frac{\sqrt[3]{4^{5-x}}}{8} = \frac{1}{2^{2x+1}},$

we can follow these steps:

Step 1: Isolate the cube root term

Multiply both sides by 8 to eliminate the denominator on the left-hand side:

$\sqrt[3]{4^{5-x}} = \frac{8}{2^{2x+1}}.$

Step 2: Simplify the right-hand side

Notice that $8$ can be rewritten as $2^3$:

$\sqrt[3]{4^{5-x}} = \frac{2^3}{2^{2x+1}} = 2^{3 - (2x+1)} = 2^{2 - 2x}.$

Step 3: Rewrite the left-hand side

Recall that $4 = 2^2$, so we can express $4^{5-x}$ as:

[ \sqrt[3]{4^{5-x}} = \sqrt[3]{(2^2)^{5-x}} = \sqrt[3]{2^{2(5-x)}} = 2^{\frac{2(5-x)}{3}}. ]

Step 4: Equate the exponents

Now, equate the exponents of $2$ on both sides:

$\frac{2(5-x)}{3} = 2 - 2x.$

Step 5: Solve for $x$

Multiply both sides by 3 to clear the fraction:

$2(5-x) = 3(2 - 2x).$

Distribute on both sides:

$10 - 2x = 6 - 6x.$

Add $6x$ to both sides:

$10 + 4x = 6.$

Subtract 10 from both sides:

$4x = -4.$

Finally, divide by 4:

$x = -1.$

So, the solution is $x = -1$.

Would you like further details, or do you have any questions?

Related Questions

1. What would the solution be if the equation was $\frac{\sqrt[3]{4^{5-x}}}{8} = \frac{1}{4^{2x+1}}$?
2. How does changing the exponent in the original equation affect the solution?
3. Can this equation be solved graphically, and what would that involve?
4. What happens if you change the cube root to a square root?
5. How does the base of the exponents influence the complexity of the equation?
6. What if the constant on the right-hand side was different, say $\frac{1}{3^{2x+1}}$?
7. What other methods can be used to solve this type of exponential equation?
8. How would the equation change if the left-hand side had a different coefficient?

Tip

When solving exponential equations, it's often helpful to express all terms with the same base. This can simplify the comparison of exponents and lead to quicker solutions.