Question:

**Step 1: Identify the fourth term using binomial expansion.** The general term in the binomial expansion of $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. For the expansion of $\left(\dfrac{2}{x} + x^{\log_8 x}\right)^6$, we have $n=6$, $a=\dfrac{2}{x}$, and $b=x^{\log_8 x}$. For the fourth term, we set $r=3$: $$T_4 = \binom{6}{3} \left(\frac{2}{x}\right)^{6-3} \left(x^{\log_8 x}\right)^3$$ $$T_4 = 20 \left(\frac{2}{x}\right)^3 \left(x^{\log_8 x}\right)^3$$ **Step 2: Simplify the fourth term.** $$T_4 = 20 \cdot \frac{8}{x^3} \cdot x^{3\log_8 x}$$ $$T_4 = 160 \cdot x^{-3} \cdot x^{3\log_8 x}$$ $$T_4 = 160 \cdot x^{3\log_8 x - 3}$$ **Step 3: Simplify the exponent using logarithm properties.** Let $y = \log_8 x$. From the definition of logarithms, this implies $x = 8^y$. Substitute $x=8^y$ into the exponent expression: $$x^{3\log_8 x - 3} = (8^y)^{3y - 3}$$ $$x^{3\log_8 x - 3} = 8^{y(3y - 3)}$$ $$x^{3\log_8 x - 3} = 8^{3y^2 - 3y}$$ **Step 4: Set up the equation using the given condition.** The fourth term is given as $20 \times 8^7$. Equating this with our simplified expression for $T_4$: $$160 \cdot 8^{3y^2 - 3y} = 20 \times 8^7$$ Divide both sides by 20: $$8 \cdot 8^{3y^2 - 3y} = 8^7$$ Using the property $a^m \cdot a^n = a^{m+n}$: $$8^{1 + 3y^2 - 3y} = 8^7$$ **Step 5: Equate exponents and solve for $y$.** Since the bases are equal, the exponents must be equal: $$1 + 3y^2 - 3y = 7$$ Rearrange into a standard quadratic form: $$3y^2 - 3y - 6 = 0$$ Divide the entire equation by 3: $$y^2 - y - 2 = 0$$ Factor the quadratic equation: $$(y - 2)(y + 1) = 0$$ This yields two possible values for $y$: $$y = 2 \quad \text{or} \quad y = -1$$ **Step 6: Convert back to $x$.** Recall that $y = \log_8 x$. Case 1: $y = 2$ $$\log_8 x = 2$$ $$x = 8^2$$ $$x = 64$$ Case 2: $y = -1$ $$\log_8 x = -1$$ $$x = 8^{-1}$$ $$x = \frac{1}{8}$$ **Step 7: Verify the solutions.** We verify if these values of $x$ satisfy the given condition. For $x=64$: If $x=64$, then $y=\log_8 64 = 2$. Substituting $y=2$ into the exponent $1 + 3y^2 - 3y$: $1 + 3(2)^2 - 3(2) = 1 + 3(4) - 6 = 1 + 12 - 6 = 7$. So, $8^{1 + 3y^2 - 3y} = 8^7$. This matches the right side of the equation $8^{1 + 3y^2 - 3y} = 8^7$, confirming $x=64$ is a valid solution. For $x=1/8$: If $x=1/8$, then $y=\log_8 (1/8) = -1$. Substituting $y=-1$ into the exponent $1 + 3y^2 - 3y$: $1 + 3(-1)^2 - 3(-1) = 1 + 3(1) + 3 = 1 + 3 + 3 = 7$. So, $8^{1 + 3y^2 - 3y} = 8^7$. This also matches the right side of the equation, confirming $x=1/8$ is a valid solution. Both $x=64$ and $x=1/8$ are positive values, satisfying the condition $x>0$. A value of $x$ is $64$.