14. (a) Evaluate $ ext{log}_2 25$ - Scottish Highers Maths - Question 14 - 2016

From part (a), we found that $ext{log}_2 25 ext{ is approximately } 4.64$. To solve $ext{log}_4 (x + ext{log}_4 (x - 6)) = 4.64$, we first convert the left-hand side using the change of base formula: ext{log}_4 (x + ext{log}_4 (x - 6)) = rac{ ext{log}_2 (x + ext{log}_4 (x - 6))}{ ext{log}_2 4} = rac{1}{2} ext{log}_2 (x + ext{log}_4 (x - 6)) Setting this equal to 4.64, we can rearrange to find: $ext{log}_2 (x + ext{log}_4 (x - 6)) = 9.28$ Taking the antilogarithm gives: $x + ext{log}_4 (x - 6) = 2^{9.28}$ We simplify this by expressing $ext{log}_4 (x - 6)$ as: ext{log}_4 (x - 6) = rac{ ext{log}_2 (x - 6)}{2} Thus, we have: x + rac{ ext{log}_2 (x - 6)}{2} = 2^{9.28} We can rewrite this expression to form a quadratic equation in $x$ and solve for $x$, identifying the appropriate solution such that $x > 6$.