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drill 2 answers can be found in part iv. a. (3^{3}\times3^{2}=) b. (\fr…

Question

drill 2
answers can be found in part iv.
a. (3^{3}\times3^{2}=)
b. (\frac{3^{3}}{3^{2}}=)
c. ((3^{3})^{2}=)
d. (x^{6}\times x^{2}=)
e. (\frac{x^{6}}{x^{2}}=)
f. ((x^{6})^{2}=)
g. (sqrt{8}=)
h. (sqrt3{-64}=)
i. (sqrt{12}+5sqrt{3}=)
j. (sqrt{y^{3}}=)
k. (sqrt3{-y^{3}}=)
l. (sqrt{x^{2}y}+5xsqrt{y}=)
1 mark for review
if (3^{4}=9^{x}), what is the value of (x)?
a 2
b 3
c 4
d 5
2 mark for review
if ((3^{x})^{3}=3^{15}), what is the value of (x)?
a 3
b 5
c 7
d 9

Explanation:

Step1: Solve a.

Use the rule $a^m\times a^n=a^{m + n}$, so $3^3\times3^2=3^{3 + 2}=3^5 = 243$.

Step2: Solve b.

Use the rule $\frac{a^m}{a^n}=a^{m - n}$, so $\frac{3^3}{3^2}=3^{3-2}=3^1 = 3$.

Step3: Solve c.

Use the rule $(a^m)^n=a^{mn}$, so $(3^3)^2=3^{3\times2}=3^6=729$.

Step4: Solve d.

Use the rule $a^m\times a^n=a^{m + n}$, so $x^6\times x^2=x^{6 + 2}=x^8$.

Step5: Solve e.

Use the rule $\frac{a^m}{a^n}=a^{m - n}$, so $\frac{x^6}{x^2}=x^{6 - 2}=x^4$.

Step6: Solve f.

Use the rule $(a^m)^n=a^{mn}$, so $(x^6)^2=x^{6\times2}=x^{12}$.

Step7: Solve g.

Simplify $\sqrt{8}=\sqrt{4\times2}=2\sqrt{2}$.

Step8: Solve h.

$\sqrt[3]{-64}=\sqrt[3]{(-4)^3}=-4$.

Step9: Solve i.

Simplify $\sqrt{12}=\sqrt{4\times3}=2\sqrt{3}$, then $\sqrt{12}+5\sqrt{3}=2\sqrt{3}+5\sqrt{3}=7\sqrt{3}$.

Step10: Solve j.

$\sqrt{y^3}=\sqrt{y^2\times y}=y\sqrt{y}$.

Step11: Solve k.

$\sqrt[3]{-y^3}=-y$.

Step12: Solve l.

$\sqrt{x^2y}=|x|\sqrt{y}$, so $\sqrt{x^2y}+5x\sqrt{y}=|x|\sqrt{y}+5x\sqrt{y}$. If $x\geq0$, it is $x\sqrt{y}+5x\sqrt{y}=6x\sqrt{y}$; if $x<0$, it is $-x\sqrt{y}+5x\sqrt{y}=4x\sqrt{y}$.

Step13: Solve 1.

Rewrite $9$ as $3^2$, so the equation $3^4 = 9^x$ becomes $3^4=(3^2)^x$. Using the rule $(a^m)^n=a^{mn}$, we have $3^4 = 3^{2x}$. Then $4 = 2x$, and $x = 2$.

Step14: Solve 2.

Use the rule $(a^m)^n=a^{mn}$, so $(3^x)^3=3^{3x}$. The equation $(3^x)^3 = 3^{15}$ gives $3x=15$, and $x = 5$.

Answer:

a. $243$
b. $3$
c. $729$
d. $x^8$
e. $x^4$
f. $x^{12}$
g. $2\sqrt{2}$
h. $-4$
i. $7\sqrt{3}$
j. $y\sqrt{y}$
k. $-y$
l. If $x\geq0$, $6x\sqrt{y}$; if $x<0$, $4x\sqrt{y}$

  1. A. $2$
  2. B. $5$