# TSI Math FREE Sample Practice Questions

Preparing for the TSI Math test? To do your best on the TSI Math test, you need to review and practice real TSI Math questions.  There’s nothing like working on TSI Math sample questions to hone your math skills and put you more at ease when taking the TSI Math test. The sample math questions you’ll find here are brief samples designed to give you the insights you need to be as prepared as possible for your TSI Math test.

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Start preparing for the 2022 TSI Math test with our free sample practice questions. Also, make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions you need to practice.

$16.99 Satisfied 105 Students ## 10 Sample TSI Math Practice Questions 1- Solve the equation: $$log_{4⁡}(x+2) – log_4⁡(x-2) = 1$$ A. $$10$$ B. $$\frac{3}{10}$$ C. $$\frac{10}{3}$$ D. $$3$$ 2- Solve: $$e^{(5x + 1 )}= 10$$ A. $$\frac{ln⁡(10) + 1}{5}$$ B. $$\frac{ln⁡(10) – 1}{5}$$ C. $$5ln (10) + 2$$ D. $$5ln (10) – 2$$ 3- If $$f(x) = x – \frac{5}{3}$$ and $$f –1$$ is the inverse of $$f(x)$$, what is the value of $$f –1(5)$$? A. $$\frac{10}{3}$$ B. $$\frac{3}{20}$$ C. $$\frac{20}{3}$$ D. $$\frac{3}{10}$$ 4- What is cos 30$$^\circ$$? A. $$\frac{1}{2}$$ B. $$\frac{{\sqrt{2}}}{2}$$ C. $$\frac{{\sqrt{3}}}{2}$$ D. $$\sqrt{3}$$ 5- If $$\theta$$ is an acute angle and sin $$\theta$$ = $$\frac{3}{5}$$, then cos $$\theta$$ = ? A. $$-1$$ B. $$0$$ C. $$\frac{4}{5}$$ D. $$\frac{5}{4}$$ 6- What is the solution of the following system of equations? $$-2x- y = -9$$ $$5x-2y= 18$$ A. $$(–1, 2)$$ B. $$(4, 1)$$ C. $$(1, 4)$$ D. $$(4, -2)$$ 7- Solve. $$|9 – (12 ÷ | 2 – 5 |)| = ?$$ A. $$9$$ B. $$-6$$ C. $$5$$ D. $$-5$$ 8- If $$log_2⁡x = 5$$, then $$x =$$ ? A. $$2^{10}$$ B. $$\frac{5}{2}$$ C. $$2^{6}$$ D. $$32$$ 9- What’s the reciprocal of $$\frac{x^3}{16}$$ ? A. $$\frac{16}{x^3}-1$$ B. $$\frac{48}{x^3}$$ C. $$\frac{16}{x^3}+1$$ D. $$\frac{16}{x^3}$$ 10- Find the inverse function for $$ln (2x + 1)$$ A. $$\frac{1}{2}(e^{x }– 1)$$ B. $$(e^{x }+ 1)$$ C. $$\frac{1}{2}(e^{x }+ 1)$$ D. $$(e^{x }– 1)$$ ## Best TSI Math Prep Resource for 2022$29.99
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1- C
METHOD ONE
$$log_4⁡(x+2) – log_4⁡(x-2) = 1$$
Add$$\space log_4⁡(x-2)$$to both sides
$$log_4⁡(x+2) – log_4⁡(x-2)+ log_4⁡(x-2)= 1 + log_4⁡(x-2)$$
$$log_4⁡(x+2) = 1 + log_4⁡(x-2)$$
Apply logarithm rule:
$$a = log_b⁡(b^a) ⇒ 1 = log_4⁡(4^1) = log_4⁡(4)$$
$$then: log_4⁡(x+2) = log_4⁡(4) + log_4⁡(x-2)$$
$$Logarithm \space rule: log_c⁡(a) + log_c⁡(b) = log_c⁡(ab)$$
$$then: log_4⁡(4) + log_4⁡(x-2) = log_4⁡(4(x-2))$$
$$log_4⁡(x+2) = log_4⁡(4(x-2))$$
When the logs have the same base:
$$log_b⁡(f(x)) = log_b⁡(g(x)) ⇒ f(x) = g(x) (x+2) = 4(x-2)$$
$$x = \frac{10}{3}$$
METHOD TWO
We know that:
$$log_a⁡b-log_a⁡c=log_a\frac{b}{c}⁡\space and \space log_a⁡b=c⇒b=a^c$$
Then: $$log_4⁡(x+2)- log_4⁡(x-2)=log_4\frac{(x + 2)}{(x – 2)}⁡=1⇒\frac{(x + 2)}{(x-2)}=4^1=4⇒x+2=4(x-2)⇒x+2=4x-8⇒4x-x=8+2→3x=10⇒x=\frac{10}{3}$$

2- B
$$e^{(5x + 1 )}= 10$$
If$$\space f(x) = g(x)$$, then $$\space ln(f(x)) = ln(g(x))$$
$$ln⁡(e^{(5x + 1 )})= ln(10)$$
Apply logarithm rule:
$$log_a⁡(x^b) = b log_a⁡(x)$$
$$ln⁡(e^{(5x + 1 )})= (5x + 1)ln(e)$$
$$(5x + 1)ln(e) = ln(10)$$
$$(5x + 1)ln(e) = (5x + 1)$$
$$(5x + 1) = ln(10)$$
$$⇒x = \frac{ln⁡(10) – 1}{5}$$

3- C
$$f(x) = x –\frac{5}{3}⇒ y = x – \frac{5}{3}⇒ y+ \frac{5}{3}=x$$
$$f^{-1 }= x+ \frac{5}{3}$$
$$f ^{–1}(5) = 5 +\frac{5}{3}=\frac{20}{3}$$

4- C
$$cos$$$$30^\circ = \frac{\sqrt 3}{2}$$

5- C

sin$$θ=\frac{3}{5}$$⇒ we have following triangle, then:
$$c=\sqrt {(5^2-3^2 )}=\sqrt{(25-9)}=\sqrt 16=4$$
cos$$θ=\frac{4}{5}$$

6- B
$$-2x- y = -9$$
$$5x-2y= 18$$
⇒ Multiplication $$(–2)$$ in first equation
$$4x +2y =18$$
$$5x-2y= 18$$
Add two equations together ⇒$$9x =36 ⇒ x= 4$$ then: $$y = 1$$

7- C
$$|9 – (12 ÷ | 2 – 5 |)| = |(9-(12÷|-3|))|=|9-(12÷3)|=|9-4|=|5|=5$$

8- D
METHOD ONE
$$log_2⁡x = 5$$
Apply logarithm rule:$$a = log_b⁡(b^a)$$
$$5 = log_2⁡(2^5) = log_2⁡(32)$$
$$log_2⁡x = log_2⁡(32)$$
When the logs have the same base: $$log_b⁡(f(x)) = log_b⁡(g(x))⇒ f(x) = g(x)$$
then: $$x = 32$$
METHOD TWO
We know that:
$$log_a⁡b=c⇒b=a^c$$
$$log_2⁡x=5⇒x=2^5=32$$

9- D
$$\frac{x^3}{16}$$
⇒ reciprocal is : $$\frac{16}{x^3}$$

10- A
$$f(x) = ln (2x + 1)$$
$$y = ln (2x + 1)$$
Change variables $$x$$ and $$y: x = ln (2y + 1)$$
solve: $$x = ln (2y + 1)$$
$$y = \frac{e^{x}-1}{2}=\frac{1}{2}(e^{x} – 1)$$

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