1. State the problem: Solve the equation $\log_2 t = \frac{1}{4}t - 5$ for $t$.\n\n2. Rewrite the equation: We want to find $t$ such that $\log_2 t = \frac{1}{4}t - 5$. This means the logarithm base 2 of $t$ equals $\frac{1}{4}t - 5$.\n\n3. Check the domain: Since $\log_2 t$ is defined only for $t > 0$, the solution must satisfy $t > 0$.\n\n4. Express the equation in exponential form: To work with the logarithmic term, rewrite the equation by isolating $\log_2 t$. It implies $t = 2^{\frac{1}{4}t - 5}$.\n\n5. Set up the equation: $$t = 2^{\frac{1}{4}t - 5} = 2^{\frac{t}{4}} \times 2^{-5} = \frac{2^{\frac{t}{4}}}{32}$$\n\n6. Multiply both sides of the equation by 32 to clear the denominator: $$32 t = 2^{\frac{t}{4}}$$\n\n7. Introduce substitution: let $x = \frac{t}{4}$, thus $t = 4x$. Substitute into the equation: $$32 \times 4x = 2^x \Rightarrow 128 x = 2^x$$\n\n8. Solve $128 x = 2^x$: This transcendental equation cannot be solved exactly by elementary algebra. We can attempt to find approximate solutions graphically or numerically.\n\n9. Analyze approximate values:\n- For $x=5$, Left side: $128 \times 5 = 640$, Right side: $2^5=32$ (LHS > RHS)\n- For $x=8$, Left side: $128 \times 8=1024$, Right side: $2^8=256$ (LHS > RHS)\n- For $x=10$, Left side: $128 \times 10 = 1280$, Right side: $2^{10} = 1024$ (LHS > RHS)\n- For $x=11$, Left side: $128 \times 11=1408$, Right side: $2^{11}=2048$ (RHS > LHS)\n\nBetween $10$ and $11$, $2^x$ surpasses $128x$, so the solution for $x$ is approximately between 10 and 11.\n\n10. Use a more precise numerical approximation (e.g., trial at $x=10.5$):\n- LHS: $128 \times 10.5 = 1344$\n- RHS: $2^{10.5} = 2^{10} \times 2^{0.5} = 1024 \times \sqrt{2} \approx 1024 \times 1.414 = 1448.15$\nSince RHS > LHS at $x=10.5$, solution is between $10$ and $10.5$.\n\n11. Estimate the solution to be approximately $x \approx 10.3$, so\n$$t = 4x \approx 4 \times 10.3 = 41.2$$\n\n12. Final answer: The approximate solution to the equation is $$\boxed{t \approx 41.2}$$