# MPE Practice Problems for Math 147, 151, or 171

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1. Rationalize the denominator. $$\displaystyle \frac{14}{3+\sqrt{2}}$$

\begin{align} \left(\dfrac{14}{3+\sqrt{2}}\right) \left(\dfrac{3-\sqrt{2}}{3-\sqrt{2}}\right)&=\dfrac{42-14\sqrt{2}}{9-2}\\ &=\dfrac{7\left(6-2\sqrt{2}\right)}{7}\\ &=6-2\sqrt{2} \end{align}

2. Find the sum or difference as indicated, and write your answer in simplified form.$\frac{x+2a-3}{x+a}-\frac{x+6}{2x}$

\begin{align*} \frac{x+2a-3}{x+a}-\frac{x+6}{2x} &=\left(\frac{x+2a-3}{x+a}\right) \left(\frac{2x}{2x}\right) - \left(\frac{x+6}{2x}\right)\left(\frac{x+a}{x+a}\right)\\ &=\frac{(x+2a-3)(2x)-(x+6)(x+a)}{(x+a)(2x)}\\ &=\frac{2x^2+4xa-6x-x^2-6x-xa-6a}{(x+a)(2x)}\\ &=\frac{x^2+3xa-12x-6a}{(x+a)(2x)} \end{align*}

3. Factor and reduce to simplest form.$\frac{6x^2+11xy-10y^2}{3x^2+10xy-8y^2}$

$\dfrac{6x^2+11xy-10y^2}{3x^2+10xy-8y^2} = \dfrac{(3x-2y)(2x+5y)}{(3x-2y)(x+4y)}=\dfrac{2x+5y}{x+4y}$

4. Simplify the following completely.$\frac{\left(x^{-4}y^{2/5}\right)^{-3/4}}{x^{2/3}y^{-5/6}}$

\begin{align*} \frac{\left(x^{-4}y^{2/5}\right)^{-3/4}}{x^{2/3}y^{-5/6}} &=\frac{x^{12/4}y^{-6/20}}{x^{2/3}y^{-5/6}}\\ &=x^{3-2/3}y^{-6/20-^-5/6}\\ &=x^{7/3}y^{32/60}\\ &=x^{7/3}y^{8/15}\\ \end{align*}

5. Solve the following equation: $$5(x-7)-13(x-7)-6=0$$.

\begin{align*} 5(x-7)-13(x-7)-6&=0\\ (x-7)(5-13)-6&=0\\ (x-7)(-8)-6&=0\\ -8x+56-6&=0\\ -8x+50&=0\\ -8x&=-50\\ x&=\dfrac{50}{8} \end{align*}

6. Find the point $$(x,y)$$ which satisfies both equations. What is the value of $$x+y$$?\begin{align*} -2x+4y&=12\\ 3x-5y&=-3\end{align*}

Multiply $$-2x+4y=12$$ by $$3$$ and $$3x-5y=-3$$ by 2. This gives
\begin{align*} -6x+12y&=36\\ 6x-10y&=-6\\ \end{align*}
Add these two equations and we get $$2y=30$$ $$\enspace \Longrightarrow \enspace$$ $$y=15$$. Substituting $$y=15$$ into $$3x-5y=-3$$ gives $$x=24$$. So the point that satisfies both equations is $$(24,15)$$, and the value of $$x+y$$ is $$24+15 =39$$.

7. Two investments are made, totaling $$10,000$$. In one year, these investments yield $$650$$ in simple interest. Part of the $$10,000$$ is invested at $$5\frac{1}{2}\%$$, and the rest at $$6\frac{3}{4}\%$$. How much more money is invested at $$6\frac{3}{4}\%$$?

Let $$x$$ be the amount invested at $$5\frac{1}{2}\%$$ and $$y$$ be the amount invested at $$6\frac{3}{4}\%$$. The resulting system of equations is
\begin{align*} x+y&=10,000\\ 0.055x+0.0675y&=650 \end{align*}
Solve the system of equations to find $$x$$ and $$y$$.
$$x=10,000-y$$ $$\enspace\Longrightarrow\enspace$$ $$0.055(10,000-y)+0.0675y=650$$ $$\enspace\Longrightarrow\enspace$$ $$550-0.055y+0.0675y=650$$ $$\enspace\Longrightarrow\enspace$$ $$0.0125y = 100$$  $$\enspace\Longrightarrow\enspace$$ $$y=8000$$. Since $$x+y=10,000$$, we know that $$x=2000$$, and thus $$6,000$$ more is invested at $$6\frac{3}{4}\%$$.

8. Given the linear equation $$2ax + 3by = 7c$$, where $$a$$, $$b$$ and $$c > 0$$, if $$x$$ decreases by $$10$$ units, what is the corresponding change in $$y$$?

Since $$2ax+3by=7c$$ $$\enspace\Longrightarrow\enspace$$ $$3by=7c-2ax$$ $$\enspace\Longrightarrow\enspace$$ $$y=\dfrac{7c-2ax}{3b}=\dfrac{-2a}{3b}x +\dfrac{7c}{3b}$$. If $$x$$ decreases by $$10$$, then we know $$y=\dfrac{-2a}{3b}(x-10)+\dfrac{7c}{3b}=\dfrac{-2ax}{3b}+\dfrac{20a}{3b}+\dfrac{7c}{3b}$$, so $$y$$ will increase by $$\dfrac{20a}{3b}$$.

9.  Line $$A$$ passes through the points $$(2k+3, 4k−6)$$ and $$(−2, 16)$$. Find the value of $$k$$ if line $$A$$ has a slope of $$0$$.

$$\dfrac{4k-6-16}{2k+3+2}=0$$ $$\enspace\Longrightarrow\enspace$$ $$\dfrac{4k-22}{2k+5}=0$$ $$\enspace\Longrightarrow\enspace$$ $$4k-22=0$$ $$\enspace\Longrightarrow\enspace$$ $$k=\dfrac{22}{4}=5.5$$

10. Jay wants to make a box, with no lid (or top), out of a $$10'' \times 6''$$ rectangular piece of cardboard. If Jay cuts squares with dimensions $$x$$ by $$x$$ out of each corner of the cardboard, and then folds up the corners to make an open box, find a function that represents:
1. The volume of the box.
2. The surface area of the box.

1. $$V=(10-2x)(6-2x)x=4x^3-32x^2+60x$$
2.  \begin{align*} SA&=2x(6-2x)+2x(10-2x)+(6-2x)(10-2x)\\&=12x-4x^2+20x-4x^2+60-32x+4x^2\\&=-4x^2+60\end{align*}

11. Solve for $$x$$ in the inequality $$\dfrac{5x+2}{x-10}\geq 3$$.

\begin{align*} \dfrac{5x+2}{x-10}&\geq 3\\ \dfrac{5x+2}{x-10}-3&\geq 0\\ \dfrac{5x+2-3(x-10)}{x-10} &\geq 0\\ \dfrac{5x+2-3x+30}{x-10}&\geq 0\\ \dfrac{2x+32}{x-10}&\geq 0 \end{align*}
The expression $$\dfrac{2x+32}{x-10}$$ will be greater than or equal to zero on $$(-\infty, -16]\cup(10,\infty)$$.

12. Find the domain of the function below.$f(x)=\dfrac{\sqrt{x^2-3x-4}}{6x^2-54}$

To find the domain of $$f(x)=\dfrac{\sqrt{x^2-3x-4}}{6x^2-54}$$ we know $$x^2-3x-4\geq 0$$ and $$6x^2-54 \neq 0$$. $$x^2-3x-4\geq 0$$ for $$(-\infty, -1]\cup[4,\infty)$$ and $$6x^2-54\neq 0$$ for $$(-\infty,-3)\cup(-3,3)\cup(3,\infty)$$. So the domain of $$f(x)$$ is $$(-\infty,-3)\cup (-3,-1]\cup [4,\infty).$$

13. Find the domain of the function below. $f(x)=\left\{\begin{array}{cc}\dfrac{2x^2+13}{x^2-1}, & x<0\\ \dfrac{5x-26}{x+2}, & x\geq 0\end{array} \right.$

For the function $$\dfrac{2x^2+13}{(x+1)(x-1)}$$ on $$x<0$$ the domain is $$(-\infty,-1)\cup(-1,0)$$. For the function $$\dfrac{5x-26}{x+2}$$ on $$x\geq 0$$ the domain is $$[0,\infty)$$. So for the function $$f(x)$$, the domain is $$(-\infty,-1)\cup (-1,\infty)$$.

14. Find the $$x$$-intercept(s) of the function $$f(x)=\dfrac{6x^2-7x-5}{4x^2-12x-7}$$, if any exist.

\begin{align*} f(x)&=\dfrac{6x^2-7x-5}{4x^2-12x-7}\\ &=\dfrac{(2x+1)(3x-5)}{(2x+1)(2x-7)}\\ &=\dfrac{3x-5}{2x-7} \end{align*}
The $$x$$-intercept is where $$3x-5=0$$ $$\enspace\Longrightarrow\enspace$$ $$x=\dfrac{5}{3}$$.

15. Find the vertical and horizontal asymptote(s) of the function $$f(x)=\dfrac{6x^2-7x-5}{4x^2-12x-7}$$, if any exist.

\begin{align*} f(x)&=\frac{6x^2-7x-5}{4x^2-12x-6}\\ &=\frac{(2x+1)(3x-5)}{(2x+1)(2x-7)}\\ &=\frac{3x-5}{2x-7} \end{align*}
The vertical asymptote is where $$2x-7=0$$ so it is $$x=\dfrac{7}{2}$$.

The horizontal asymptote is at $$y= \dfrac{6}{4}$$ which reduces to $$y= \dfrac{3}{2}$$.

16. Find the $$x$$- and $$y$$-intercepts for the function $$f(x)=x^3-9x$$.

To find the $$x$$-intercepts solve $$f(x)=0$$ $$\enspace \Longrightarrow\enspace$$ $$x(x+3)(x-3)=0$$ $$\enspace \Longrightarrow\enspace$$ $$x=0,\pm 3$$.

To find the $$y$$-intercept, substitute $$x=0$$ $$\enspace \Longrightarrow\enspace$$ $$0^3-9(0)=0$$. Hence, the $$y$$-intercept is $$y=0$$.

17. Find the domain of the following functions:
1. $$f(x)=\sqrt{-x^2-4x+5}$$
2. $$g(t)=\ln(4t-3)$$
3. $$h(x)=\dfrac{1}{x^3+3x^2-x-3}$$

1. We must have $$−x^2−4x+5\geq 0$$ for $$f(x)$$ to be defined. Thus, $$(−x+1)(x+5)\geq 0$$ which occurs on $$[−5, 1]$$. Thus, the domain of $$f(x)$$ is $$[−5, 1]$$.
2. We must have $$4t − 3 > 0$$ for $$g(t)$$ to be defined. Thus, the domain for $$g(t)$$ is $$\left(\dfrac{3}{4},\infty\right)$$.
3. We must have $$x^3+3x^2-x-3\neq 0$$ for $$h(x)$$ to be defined. \begin{align*} x^3+3x^2-x-3&\neq 0\\ x^2(x+3)-(x+3)&\neq 0\\ (x+3)\left(x^2-1\right) &\neq 0 \x+3)(x-1)(x+1)&\neq 0 \end{align*} So \(x\neq \pm 1 and $$x\neq -3$$. Thus, the domain of $$h(x)$$ is $$(-\infty, -3)\cup (-3,-1) \cup (-1,1) \cup (1,\infty)$$.

18. Simplify the expression $$\dfrac{2}{\sqrt{x^5}} \left(\sqrt[3]{4x}\right)$$.

\begin{align*} \left( \dfrac{2}{\sqrt{x^5}}\right) \left(\sqrt[3]{4x}\right) &= \left(2x^{-5/2} \right) (4x)^{1/3}\\ &=\left(2x^{-5/2}\right) \left(4^{1/3}\cdot x^{1/3}\right) \\ &=\left(2x^{-5/2}\right)\left( \left(2^2\right)^{1/3}x^{1/3}\right)\\ &=\left(2x^{-5/2}\right)\left(2^{2/3}x^{1/3}\right)\\ &=\dfrac{2^{5/3}}{x^{13/6}} \end{align*}

19. If we begin with the graph of $$f(x) = x^2$$ and shift $$f (x)$$ 4 units to the right, shrink f (x) vertically by a factor of $$\frac{1}{2}$$, and then shift $$f(x)$$ upward 10 units, write the equation for the transformed graph.

The transformed graph is $$g(x)=\dfrac{1}{2}(x-4)^2+10$$.

20. Solve the following equation for $$x$$.$\log(x+2)+\log(x-1)=1$

The domain of $$\log(x+2)$$ is $$(-2,\infty)$$. The domain of $$\log(x-1)$$ is $$(1,\infty)$$. So, the domain of $$\log(x+2)+\log(x-1)$$ is $$(1,\infty)$$. Now, use properties of logarithms to solve the equation.
\begin{align*} \log(x+2)+\log(x-1)&=1\\ \log\left[ (x+2)(x-1)\right] &= 1\\ (x+2)(x-1)&=10^1\\ x^2+x-2&=10\\ x^2+x-12&=0\\ (x+4)(x-3)&=0\\ x&=-4,3 \end{align*}
Since $$x=-4$$ is not in the domain, $$x=3$$ is the only solution.

21. Factor the expression below completely. $3x^2\left(4x^2+1\right)^8+64x^4\left(4x^2+1\right)^7$

\begin{align*} 3x^2\left(4x^2+1\right)^8+64x^4\left(4x^2+1\right)^7 &= x^2\left(4x^2+1\right)^7 \left(3\left(4x^2+1\right)+64x^2\right)\\ &=x^2\left(4x^2+1\right)^7\left(12x^2+3+64x^2\right)\\ &=x^2\left(4x^2+1\right)^7\left(76x^2+3\right) \end{align*}

22. How far from the base of an $$18$$ foot tall pole must a person be standing if the angle of elevation from the ground to the pole is $$41^\circ$$?

Refer to the figure below. We know that $$\tan (41^\circ)=\dfrac{18}{x}$$. So, $$x=\dfrac{18}{\tan 41^\circ}$$ or $$x=18\cot(41^\circ)$$. So, the person must be $$18\cot(41^\circ)$$ feet from the base of the pole.

23. Find $$f\circ g$$ (also denoted $$f(g(x))$$) if $$f(x)=\dfrac{x}{x+1}$$ and $$g(x)=\dfrac{2}{x}$$. Simplify.

\begin{align*} f\circ g &= \frac{ \frac{2}{x}}{\frac{2}{x}+1}\\ &= \frac{ \frac{2}{x}}{\frac{2}{x}+\frac{x}{x}}\\ &=\frac{ \frac{2}{x}}{\frac{2+x}{x}}\\ &=\frac{2}{x}\cdot \frac{x}{2+x} \\ &=\frac{2}{2+x} \end{align*}

24. Perform the indicated operations and simplify. $\frac{8}{x+1}-\left(\frac{y}{z+2} \div \frac{y-4}{w}\right)$

\begin{align*} \frac{8}{x+1}-\left( \frac{y}{z+2} \div \frac{y-4}{w}\right) &= \frac{8}{x+1}-\left( \frac{y}{z+2} \cdot \frac{w}{y-4}\right)\\ &=\frac{8}{x+1} - \left( \frac{yw}{(z+2)(y-4)}\right) \\ &=\frac{ 8(z+2)(y-4)-yw(x+1)}{(x+1)(z+2)(y-4)}\\ &=\frac{8zy-32z+16y-64-ywx-yw}{(x+1)(z+2)(y-4)} \end{align*}

25. Solve the equation $$e^{2x}-2e^x-3=0$$ for $$x$$.

First, factor the equation: $$e^{2x}-2e^x-3=(e^x)^2-2e^x-3=(e^x-3)(e^x+1)$$. Solving $$(e^x-3)(e^x+1)=0$$ $$\enspace \Longrightarrow\enspace$$ $$e^x-3=0$$ or $$e^x+1=0$$ $$\enspace \Longrightarrow\enspace$$ $$e^x=3$$ or $$e^x=-1$$. But $$e^x$$ will never be negative, so the only solution is $$e^x=3$$ $$\enspace \Longrightarrow\enspace$$ $$x=\ln(3)$$.

26. Find the equation of the line passing through the point $$(5,1)$$ with a slope of 7. Use the equation you find to determine the value of $$y$$ when $$x=-4.$$

Using the point-slope equation, we get $$y-1=7(x-5)$$ $$\enspace \Longrightarrow\enspace$$ $$y=7x-35+1$$ $$\enspace \Longrightarrow\enspace$$ $$y=7x-34$$. Using this to find $$y$$ when $$x=-4$$ gives $$y=7(-4)-34$$ $$\enspace \Longrightarrow\enspace$$ $$y=-62$$.

27. If $$f(x)=\sqrt{x+4}$$, find and simplify $$\dfrac{f(2+h)-f(2)}{h}$$.

\begin{align*} \dfrac{f(2+h)-f(2)}{h} &=\dfrac{\sqrt{2+h+4}-\sqrt{2+4}}{h}\\ &=\left(\dfrac{\sqrt{6+h}-\sqrt{6}}{h}\right) \cdot \left( \dfrac{\sqrt{6+h}+\sqrt{6}}{\sqrt{6+h}+\sqrt{6}}\right)\\ &=\dfrac{6+h-6}{h\left(\sqrt{6+h}+\sqrt{6}\right)}\\ &=\dfrac{1}{\sqrt{6+h}+\sqrt{6}} \end{align*}

28. Simplify $$\dfrac{ \left(x^2y^4\right)^5\left( x^3y\right)^{-3}}{xy}$$.

\begin{align*} \dfrac{ \left(x^2y^4\right)^5\left( x^3y\right)^{-3}}{xy} &=\dfrac{x^{10}y^{20}x^{-9}y^{-3}}{xy}\\ &=\dfrac{xy^{17}}{xy}\\ &=y^{16} \end{align*}

29. Simplify $$\sqrt[3]{\left(a^3b\right)}\sqrt[3]{64a^4b^2}$$.

$$\sqrt[3]{\left(a^3b\right)}\sqrt[3]{64a^4b^2}=\sqrt[3]{64a^7b^3}=4a^2b\left(\sqrt[3]{a}\right)$$

30. Perform the operations indicated and simplify.$\frac{x^2}{x^2-x-2}-\frac{4}{x^2+x-6}+\frac{x}{x^2+4x+3}$

\begin{align*} \frac{x^2}{x^2-x-2}&-\frac{4}{x^2+x-6}+\frac{x}{x^2+4x+3}\\ &=\frac{x^2}{(x-2)(x+1)}-\frac{4}{(x+3)(x-2)}+\frac{x}{(x+1)(x+3)}\\ &=\left(\frac{x+3}{x+3}\right)\left(\frac{x^2}{(x-2)(x+1)}\right) -\left(\frac{x+1}{x+1}\right) \left(\frac{4}{(x+3)(x-2)}\right)+\left(\frac{x-2}{x-2}\right)\left(\frac{x}{(x+1)(x+3)}\right)\\ &= \frac{x^3+3x^2-4x-4+x^2-2x}{(x+3)(x-2)(x+1)}\\ &=\frac{x^3+4x^2-6x-4}{(x+3)(x-2)(x+1)} \end{align*}

31. Find all zeros, horizontal, and vertical asymptotes for $$f(x)=\dfrac{3x^2-14x-5}{4x^2-17x-15}$$.

Factor and simplify: $$f(x)=\dfrac{3x^2-14x-5}{4x^2-17x-15}=\dfrac{(3x+1)(x-5)}{(4x+3)(x-5)}=\dfrac{3x+1}{4x+3}$$. This yields a zero of $$x=-\dfrac{1}{3}$$, a vertical asymptote of $$x=-\dfrac{3}{4}$$, and a horizontal asymptote of $$y=\dfrac{3}{4}$$.

32. If $$\theta$$ is in Quadrant II, and $$\sin(\theta) = \dfrac{1}{7}$$, what is $$\cos(\theta)$$?

Using the identity $$\cos^2 \theta + \sin^2 \theta = 1$$, we find that $$\cos^2 \theta = 1-\sin^2\theta=1-\dfrac{1}{49}=\dfrac{48}{49}$$. This gives us $$\cos \theta= \pm \sqrt{\dfrac{48}{49}}=\pm \dfrac{\sqrt{48}}{7} = \pm \dfrac{4\sqrt{3}}{7}$$. Since $$\theta$$ is in quadrant II, $$\cos \theta <0$$, hence $$\cos \theta = -\dfrac{4\sqrt{3}}{7}$$.

33. Use the properties of logarithms to expand the expression $$\ln \left( \dfrac{\sqrt{x}y^5}{(z+1)^4}\right)$$.

We know $$\ln(ab)=\ln a + \ln b$$, $$\ln \left(\dfrac{a}{b}\right)=\ln a - \ln b$$, and $$\ln a^b =b \ln a$$. Using these properties, we obtain
\begin{align*} \ln \left( \frac{ \sqrt{x}y^5}{(z+1)^4} \right) &= \ln \left( \sqrt{x}y^5\right) - \ln (z+1)^4\\ &=\ln \left(x^{1/2}y^5\right)-4\ln(z+1)\\ &=\ln x^{1/2}+\ln y^5 -4\ln(z+1)\\ &=\dfrac{1}{2}\ln x+5\ln y -4\ln(z+1) \end{align*}

34. Evaluate $$\sec \dfrac{2\pi}{3} - \tan \dfrac{\pi}{6}$$.

$$\sec \dfrac{2\pi}{3}-\tan \dfrac{\pi}{6} = -2 - \dfrac{1}{\sqrt{3}}$$

35. If we begin with a rectangle with length 5 inches and width 4 inches, then increase the length by $$8 \%$$, what is the change in area?

We begin with a rectangle of area $$20$$ in$$^2$$. If we increase the length by $$8 \%$$, then our new length is $$5.4$$ inches. This gives a new area of $$21.6$$ in$$^2$$. This gives a total change in area of $$1.6$$ in$$^2$$.

36. Evaluate $$f(2)-f(-3)$$ given $$f(x)=\left\{ \begin{array}{ll} x^3+1, & x>1\\2x^2-3,&x\leq 1\end{array}\right.$$

$$f(2)=(2)^3+1=9$$ and $$f(-3)=2(-3)^2-3=18-3=15$$, so $$f(2)-f(-3)=9-15=-6$$.

37. Simplify the expression $$\dfrac{\cos^2 (\theta)}{1+\sin(\theta)}$$.

Use the identity $$\cos^2\theta = 1-\sin^2 \theta$$ to get
\begin{align*} \frac{\cos^2(\theta)}{1+\sin(\theta)} &= \frac{1-\sin^2\theta}{1+\sin \theta}\\ &=\frac{(1+\sin \theta)(1-\sin \theta)}{1+\sin\theta}\\ &=1-\sin \theta \end{align*}

38. Evaluate $$\log_4\left(\dfrac{1}{\sqrt[3]{16}}\right)$$.

\begin{align*} \log_4 \left(\dfrac{1}{\sqrt[3]{16}}\right) &=\log_4\left( \frac{1}{\sqrt[3]{4^2}}\right)\\ &=\log_4 \left( \dfrac{1}{4^{2/3}}\right)\\ &=\log_4 4^{-2/3} \\ &=-\dfrac{2}{3}\\ \end{align*}

39. Simplify $$\dfrac{\frac{1}{a}-b}{\frac{1}{b^3}+a}$$.

$$\dfrac{\frac{1}{a}-b}{\frac{1}{b^3}+a} = \dfrac{\left(\frac{1}{a}-b\right)\left(ab^3\right)}{\left(\frac{1}{b^3}+a\right)\left(ab^3\right)} =\dfrac{b^3-ab^4}{a+a^2b^3}$$
40. A bacteria contains $$1200$$ bacteria and doubles every day. How many hours will it take the culture to reach 10000 bacteria?
Use the exponential model $$y(t)=y_0e^{kt}$$, where $$y(t)$$ is the size of the population at time $$t$$ and $$y_0$$ is the initial size of the population. We know that $$y_0=1200$$, hence $$y(t)=1200e^{kt}$$.​ Since the population doubles everyday, we know
\begin{align*} 2400&=1200e^{k\cdot 24}\\ 2&=e^{24k}\\ \ln 2&= 24 k\\ k&=\dfrac{1}{24}\ln 2\\ k&= \ln 2^{1/24}\\ \end{align*}
Thus, $$y(t)=1200e^{t\ln 2^{1/24}} = 1200 e^{\ln 2^{t/24}}=1200\cdot 2^{t/24}$$ (note $$t$$ is in hours). Now solve for $$t$$ when $$y(t)=10000$$
\begin{align*} 10000&=1200\cdot 2^{t/24}\\ \dfrac{10000}{1200}&=2^{t/24}\\ \dfrac{25}{3}&=2^{t/24}\\ \log_2\left(\dfrac{25}{3}\right) &= \dfrac{t}{24}\\ t&=24\log_2 \left(\dfrac{25}{3}\right)\\ \end{align*}
So, it will take $$24 \log_2 \left( \dfrac{25}{3}\right)$$ hours for the culture to reach $$10,000$$ bacteria.