1. We are given the equation $\log_{2}(t) = \frac{1}{4}t - 5$. Our goal is to solve for $t$. 2. Rewrite the logarithmic equation in its equivalent exponential form. Since $\log_a(b) = c$ means $a^c = b$, we have: $$2^{\frac{1}{4}t - 5} = t$$ 3. Simplify the right side of the exponent: $$2^{\frac{t}{4} - 5} = t$$ 4. Rewrite $2^{\frac{t}{4} - 5}$ as a product of powers: $$2^{\frac{t}{4}} \times 2^{-5} = t$$ 5. Calculate $2^{-5}$: $$2^{\frac{t}{4}} \times \frac{1}{32} = t$$ 6. Multiply both sides by 32: $$2^{\frac{t}{4}} = 32t$$ 7. This is a transcendental equation and cannot be solved algebraically with elementary functions. We need to solve for $t$ numerically. 8. To solve numerically, define a function: $$f(t) = 2^{\frac{t}{4}} - 32t$$ And find $t$ such that $f(t) = 0$. 9. By testing some values: - $t=4$: $2^{1} - 128 = 2 - 128 = -126$ (negative) - $t=8$: $2^{2} - 256 = 4 - 256 = -252$ (negative) - $t=16$: $2^{4} - 512 = 16 - 512 = -496$ (negative) - $t=1$: $2^{0.25} - 32 = \approx 1.189 - 32 = -30.81$ (negative) This suggests $t$ is small. 10. Check $t=0.1$: $$2^{0.025} - 3.2 \approx 1.0175 - 3.2 = -2.18$$ (still negative) 11. Check $t=0$: $$2^{0} - 0 = 1 - 0 = 1$$ (positive) 12. Since $f(0) > 0$ and $f(0.1) < 0$, the root lies between 0 and 0.1. 13. Narrowing down numerically, approximate solution is close to $t \approx 0.03$ (numerical approximation). **Final answer:** $t \approx 0.03$ (approximate solution)