1. The problem is to solve the equation $5(3^{1-x}) = x$ for $x$. 2. We start by rewriting the equation: $$5 \cdot 3^{1-x} = x$$ 3. This is a transcendental equation because $x$ appears both as an exponent and as a linear term. 4. To solve it, we can try to isolate $x$ or use substitution, but here it is easier to test the given options to see which satisfies the equation. 5. Check option (a) $x=16/3 \approx 5.333$: $$5 \cdot 3^{1-5.333} = 5 \cdot 3^{-4.333} \approx 5 \cdot 0.011 = 0.055$$ This is not equal to $5.333$. 6. Check option (b) $x=14/3 \approx 4.667$: $$5 \cdot 3^{1-4.667} = 5 \cdot 3^{-3.667} \approx 5 \cdot 0.017 = 0.085$$ Not equal to $4.667$. 7. Check option (c) $x=42$: $$5 \cdot 3^{1-42} = 5 \cdot 3^{-41} \approx 5 \cdot \text{very small} \approx 0$$ Not equal to $42$. 8. Check option (d) $x=82/3 \approx 27.333$: $$5 \cdot 3^{1-27.333} = 5 \cdot 3^{-26.333} \approx 5 \cdot \text{very small} \approx 0$$ Not equal to $27.333$. 9. None of the options satisfy the equation exactly, but since the problem states "log 3, the value of 5 (3 1 x− ) = =x", it seems the intended solution is to find $x$ such that $5 \cdot 3^{1-x} = x$. 10. Let's try to solve it approximately by rewriting: $$x = 5 \cdot 3^{1-x}$$ 11. Taking logarithm base 3 on both sides: $$\log_3 x = \log_3 5 + 1 - x$$ 12. Rearranged: $$x + \log_3 x = 1 + \log_3 5$$ 13. Since $\log_3 5 \approx 1.46497$, then: $$x + \log_3 x \approx 2.46497$$ 14. Testing $x=2$: $$2 + \log_3 2 \approx 2 + 0.631 = 2.631 > 2.46497$$ 15. Testing $x=1.5$: $$1.5 + \log_3 1.5 \approx 1.5 + 0.369 = 1.869 < 2.46497$$ 16. So $x$ is between 1.5 and 2, closer to 2. 17. None of the given options match this approximate solution, so the problem might have a typo or is asking for the value of $x$ in terms of logarithms. Final answer: The approximate solution to $5 \cdot 3^{1-x} = x$ is about $x \approx 1.8$ (not among the options).