, 2012

Solution 1

The hour hand and minute hand both turn at constant rates. Since the hour hand moves $\frac{1}{12}$ of the way around the clock in 1 hour and the minute hand moves all of the way around the clock in 1 hour, then the minute hand turns 12 times as quickly as the hour hand.

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Suppose also that the hour hand moves through an angle of $x^\circ$ between Before and After.

Therefore, the minute hand moves through an angle of $(360^\circ - x^\circ)$ between Before and After, since these two angles add to $360^\circ$.

Since the minute hand moves 12 times as quickly as the hour hand, then $\dfrac{360^\circ - x^\circ}{x^\circ} = 12$ or $360 - x = 12x$ and so $13x = 360$, or $x = \frac{360}{13}$.

In one hour, the hour hand moves through $\frac{1}{12}\times 360^\circ = 30^\circ$.

Since the hour hand is moving for $t$ hours, then we have $30^\circ t = (\frac{360}{13})^\circ$ and so $t = \frac{360}{30(13)}=\frac{12}{13}$.

Solution 2

Suppose that Jimmy starts painting $x$ hours after 9:00 a.m. and finishes painting $y$ hours after 10:00 a.m., where $0<x<1$ and $0<y<1$.

Since $t$ is the amount of time in hours that he spends painting, then $t = (1-x)+y$, because he paints for $(1-x)$ hours until 10:00 a.m., and then for $y$ hours until his finishing time.

The hour hand and minute hand both turn at constant rates.

The minute hand turns $360^\circ$ in one hour and the hour hand turns $\frac{1}{12}\times 360^\circ = 30^\circ$ in one hour.

Thus, in $x$ hours, where $0<x<1$, the minute hand turns $(360x)^\circ$ and the hour hand turns $(30x)^\circ$.

In the Before picture, the minute hand is $(360x)^\circ$ clockwise from the 12 o’clock position.

In the After picture, the minute hand is $(360y)^\circ$ clockwise from the 12 o’clock position.

The 9 is $9\times 30^\circ = 270^\circ$ clockwise from the 12 o’clock position and the 10 is $10\times 30^\circ = 300^\circ$ clockwise from the 12 o’clock position.

Therefore, in the Before picture, the hour hand is $270^\circ + (30x)^\circ$ clockwise from the 12 o’clock position, and in the After picture, the hour hand is $300^\circ + (30y)^\circ$ clockwise from the 12 o’clock position.

Because the hour and minute hands have switched places from the Before to the After positions, then we can equate the corresponding positions to obtain $(360x)^\circ = 300^\circ + (30y)^\circ$ (or $360x=300+30y$) and $(360y)^\circ = 270^\circ + (30x)^\circ$ (or $360y = 270+30x$).

Dividing both equations by 30, we obtain $12x=10+y$ and $12y=9+x$.

Subtracting the second equation from the first, we obtain $12x-12y = 10+y-9-x$ or $-1 = 13y-13x$.

Therefore, $y-x=-\frac{1}{13}$ and so $t = (1-x)+y= 1+y-x = 1-\frac{1}{13}=\frac{12}{13}$.
We manipulate the given equation into a sequence of equivalent equations: $$\text{5x+9 (x 2 + 6x + 9) + x+3 (5x 2 + 24x + 27) = 4 (x 2 + 6x + 9) (5x+9) + (5x 2 + 24x + 27) (x+3) = 4 (using the ``change of base" formula) ((x+3) 2) (5x+9) + ((5x+9)(x+3)) (x+3) = 4 (factoring) 2 (x+3) (5x+9) + (5x+9)+ (x+3) (x+3) = 4 (using logarithm rules) 2 ( (x+3) (5x+9) ) + (5x+9) (x+3) + (x+3) (x+3) = 4 (rearranging fractions)}$$ Making the substitution $t = \dfrac{\log(x+3)}{\log(5x+9)}$, we obtain successively $$\begin{aligned} 2t + \dfrac{1}{t} + 1 & = 4 \\ 2t^2 + 1 + t & = 4t \\ 2t^2 - 3t + 1 & = 0 \\ (2t-1)(t-1) & = 0\end{aligned}$$ Therefore, $t=1$ or $t=\tfrac{1}{2}$.

If $\dfrac{\log(x+3)}{\log(5x+9)} = 1$, then $\log(x+3)=\log(5x+9)$ or $x+3=5x+9$, which gives $4x=-6$ or $x = -\frac{3}{2}$.

If $\dfrac{\log(x+3)}{\log(5x+9)} = \dfrac{1}{2}$, then $2\log(x+3)=\log(5x+9)$ or $\log((x+3)^2)=\log(5x+9)$ or $(x+3)^2=5x+9$.

Here, $x^2+6x+9=5x+9$ or $x^2+x=0$ or $x(x+1)=0$, and so $x=0$ or $x=-1$.

Therefore, there are three possible values for $x$: $x=0$, $x=-1$ and $x=-\frac{3}{2}$.

We should check each of these in the original equation.

If $x=0$, the left side of the original equation is $\log_9 9+\log_3 27 = 1 + 3 = 4$.

If $x=-1$, the left side of the original equation is $\log_4 4+\log_2 8 = 1 + 3 = 4$.

If $x=-\frac{3}{2}$, the left side of the original equation is $\log_{3/2} (9/4) + \log_{3/2} (9/4) = 2 + 2 = 4$.

Therefore, the solutions are $x=0,-1,-\frac{3}{2}$.