1. We start by solving the first logarithmic equation: $$^2\log(16) + ^2\log(x) = 6$$ 2. Recall the logarithm property: $$\log_b(a) + \log_b(c) = \log_b(ac)$$. Here, the base is 2, so we can combine the logs: $$^2\log(16) + ^2\log(x) = ^2\log(16x)$$ 3. Substitute back into the equation: $$^2\log(16x) = 6$$ 4. By definition of logarithm, this means: $$16x = 2^6$$ 5. Calculate the right side: $$2^6 = 64$$ 6. So: $$16x = 64$$ 7. Divide both sides by 16: $$x = \frac{64}{16}$$ 8. Simplify the fraction: $$x = \cancel{\frac{64}{16}} = 4$$ --- 9. Now solve the second equation: $$^2\log(2x) - \frac{1}{2} ^1\log(x) = 5$$ 10. Note that $$^1\log(x)$$ means logarithm base 1, which is undefined because logarithm base must be positive and not equal to 1. This is likely a typo or misinterpretation. 11. Assuming the second term is $$\frac{1}{2} \log_2(x)$$ (log base 2), rewrite the equation as: $$^2\log(2x) - \frac{1}{2} ^2\log(x) = 5$$ 12. Use logarithm properties: $$^2\log(2x) = ^2\log(2) + ^2\log(x) = 1 + ^2\log(x)$$ 13. Substitute into the equation: $$1 + ^2\log(x) - \frac{1}{2} ^2\log(x) = 5$$ 14. Combine like terms: $$1 + \left(1 - \frac{1}{2}\right) ^2\log(x) = 5$$ $$1 + \frac{1}{2} ^2\log(x) = 5$$ 15. Subtract 1 from both sides: $$\frac{1}{2} ^2\log(x) = 4$$ 16. Multiply both sides by 2: $$^2\log(x) = 8$$ 17. Rewrite in exponential form: $$x = 2^8$$ 18. Calculate: $$x = 256$$ Final answers: - For a: $$x = 4$$ - For b: $$x = 256$$