CEMC Euclid, 2022

We note that $\dfrac{2x+1}{x} = \dfrac{2x}{x} + \dfrac{1}{x} = 2 + \dfrac{1}{x}$.

Therefore, $\dfrac{2x+1}{x} = 4$
exactly when $2 + \dfrac{1}{x} = 4$
or $\dfrac{1}{x} = 2$ and so $x = \dfrac{1}{2}$.

Alternatively, we could solve $\dfrac{2x+1}{x} = 4$ directly to obtain
$2x + 1 = 4x$, which gives $2x = 1$ and so $x = \dfrac{1}{2}$.

Thus, to determine the value of $f(4)$, we substitute $x = \dfrac{1}{2}$ into the given equation
$f\left(\dfrac{2x+1}{x}\right) = x + 6$ and obtain $f(4) = \dfrac{1}{2} + 6 = \dfrac{13}{2}$.
Since the graph passes through $(3,5)$, $(5,4)$ and $(11,3)$, we can substitute these three
points and obtain the following three equations: $$\begin{aligned} 5 & = \log_a(3+b) + c \\ 4 & = \log_a(5+b) + c \\ 3 & = \log_a(11+b) + c \end{aligned}$$ Subtracting the second
equation from the first and the third equation from the second, we
obtain: $$\begin{aligned} 1 & = \log_a(3+b) - \log_a(5+b) \\ 1 & = \log_a(5+b) - \log_a(11+b)\end{aligned}$$ Equating
right sides and manipulating, we obtain the following equivalent
equations: $$\begin{aligned} \log_a(5+b) - \log_a(11+b) & = \log_a(3+b) - \log_a(5+b) \\ 2\log_a(5+b) & = \log_a(3+b) + \log_a(11+b) \\ \log_a\left((5+b)^2\right) & = \log_a\left((3+b)(11+b)\right) & \text{(using log laws)} \\ (5+b)^2 & = (3+b)(11+b) & \text{(raising both sides to the power of $a$)} \\ 25+10b + b^2 & = 33 + 14b + b^2 \\ -8 & = 4b \\ b & = -2\end{aligned}$$ Since $b = -2$, the equation $1 = \log_a(3+b) - \log_a(5+b)$ becomes $1 = \log_a 1 - \log_a 3$.

Since $\log_a 1 = 0$ for every
admissible value of $a$, then $\log_a 3 = -1$ which gives $a = 3^{-1} = \frac{1}{3}$.

Finally, the equation $5 = \log_a(3+b) + c$ becomes $5 = \log_{1/3}(1) + c$ and so $c = 5$.

Therefore, $a = \frac{1}{3}$, $b = -2$, and $c = 5$, which gives $y = \log_{1/3}(x-2) + 5$.

Checking:

When $x = 3$, we obtain $y = \log_{1/3}(3-2) + 5 = \log_{1/3} 1 + 5 = 0 + 5 = 5$.
When $x = 5$, we obtain $y = \log_{1/3}(5-2) + 5 = \log_{1/3} 3 + 5 = -1 + 5 = 4$.
When $x = 11$, we obtain $y = \log_{1/3}(11-2) + 5 = \log_{1/3} 9 + 5 = -2 + 5 = 3$.