QUESTION IMAGE
Question
all answers need an exponent! no exponents, no points.
guided practice
lets try these together.
- $(4^{2})^{3}$
- $(b^{7})^{8}$
- $sqrt{4^{2}}$
- $sqrt{x^{3}}$
- $sqrt3{2^{4}}$
- $sqrt3{x^{5}}$
- $sqrt3{s^{2}m^{4}h^{3} \cdot s^{7}m^{2}}$
secret parentheses!
always simplify inside the root before dividing the exponents!
independent practice
try these on your own.
- $((-2)^{3})^{3}$
- $(2^{2})^{-2}$
- $(x^{2})^{-3}$
- $sqrt{\frac{x^{6}}{x^{4}}}$
- $sqrt3{a^{2}b^{2}c^{3} \cdot ab}$
- $(3 \cdot 3^{2})^{2}$
- $(x^{6})^{5}$
- $sqrt3{\frac{2^{7}}{2^{3}}}$
- $sqrt{xy^{2}z^{4} \cdot x^{2}y^{2}}$
- $sqrt3{(z^{2})^{4}}$
- $(y^{2})^{39}$
- $(4^{2})^{3}$
- $\left(\frac{6^{4}}{6^{2}}\
ight)^{3}$
- $sqrt{\frac{x^{4}y^{2}}{x^{2}y}}$
- $sqrt{s^{2}m^{2}h^{2}}$
- $sqrt3{2^{6}}$
- $sqrt{3^{5} \cdot 3^{2}}$
- $((x^{3})^{2})^{2}$
- $sqrt3{\frac{x^{8}y^{9}}{x^{3}y^{4}}}$
- $sqrt{(x^{2})^{3}}$
Guided Practice
Step1: Power of a power rule
For $(4^2)^3$, multiply exponents: $2 \times 3 = 6$, so $4^{2 \times 3} = 4^6$
Step2: Power of a power rule
For $(b^7)^8$, multiply exponents: $7 \times 8 = 56$, so $b^{7 \times 8} = b^{56}$
Step3: Root as fractional exponent
$\sqrt{4^2} = (4^2)^{\frac{1}{2}}$, multiply exponents: $2 \times \frac{1}{2} = 1$, so $4^{2 \times \frac{1}{2}} = 4^1$
Step4: Root as fractional exponent
$\sqrt{x^3} = (x^3)^{\frac{1}{2}}$, multiply exponents: $3 \times \frac{1}{2} = \frac{3}{2}$, so $x^{3 \times \frac{1}{2}} = x^{\frac{3}{2}}$
Step5: Root as fractional exponent
$\sqrt[3]{2^4} = (2^4)^{\frac{1}{3}}$, multiply exponents: $4 \times \frac{1}{3} = \frac{4}{3}$, so $2^{4 \times \frac{1}{3}} = 2^{\frac{4}{3}}$
Step6: Root as fractional exponent
$\sqrt[3]{x^5} = (x^5)^{\frac{1}{3}}$, multiply exponents: $5 \times \frac{1}{3} = \frac{5}{3}$, so $x^{5 \times \frac{1}{3}} = x^{\frac{5}{3}}$
Step7: Simplify inside root first
Multiply like bases: $s^{2+1}m^{4+2}h^{3} = s^3m^6h^3$. Apply cube root: $(s^3m^6h^3)^{\frac{1}{3}}$, multiply exponents: $s^{3 \times \frac{1}{3}}m^{6 \times \frac{1}{3}}h^{3 \times \frac{1}{3}} = s^1m^2h^1$
Independent Practice (1-10)
Step1: Power of a power rule
For $((-2)^3)^3$, multiply exponents: $3 \times 3 = 9$, so $(-2)^{3 \times 3} = (-2)^9$
Step2: Power of a power rule
For $(2^2)^{-2}$, multiply exponents: $2 \times (-2) = -4$, so $2^{2 \times (-2)} = 2^{-4}$
Step3: Power of a power rule
For $(x^2)^{-3}$, multiply exponents: $2 \times (-3) = -6$, so $x^{2 \times (-3)} = x^{-6}$
Step4: Simplify inside root first
Subtract exponents: $\frac{x^6}{x^4} = x^{6-4} = x^2$. Apply square root: $(x^2)^{\frac{1}{2}} = x^{2 \times \frac{1}{2}} = x^1$
Step5: Simplify inside root first
Multiply like bases: $a^{2+1}b^{2+1}c^3 = a^3b^3c^3$. Apply cube root: $(a^3b^3c^3)^{\frac{1}{3}} = a^{3 \times \frac{1}{3}}b^{3 \times \frac{1}{3}}c^{3 \times \frac{1}{3}} = a^1b^1c^1$
Step6: Simplify inside parentheses first
$3 \times 3^2 = 3^{1+2} = 3^3$. Apply power of a power: $(3^3)^2 = 3^{3 \times 2} = 3^6$
Step7: Power of a power rule
For $(x^6)^5$, multiply exponents: $6 \times 5 = 30$, so $x^{6 \times 5} = x^{30}$
Step8: Simplify inside root first
Subtract exponents: $\frac{2^7}{2^3} = 2^{7-3} = 2^4$. Apply cube root: $(2^4)^{\frac{1}{3}} = 2^{4 \times \frac{1}{3}} = 2^{\frac{4}{3}}$
Step9: Simplify inside root first
Multiply like bases: $x^{1+2}y^{2+2}z^4 = x^3y^4z^4$. Apply square root: $(x^3y^4z^4)^{\frac{1}{2}} = x^{3 \times \frac{1}{2}}y^{4 \times \frac{1}{2}}z^{4 \times \frac{1}{2}} = x^{\frac{3}{2}}y^2z^2$
Step10: Power of a power first
$(z^2)^4 = z^{8}$. Apply cube root: $(z^8)^{\frac{1}{3}} = z^{8 \times \frac{1}{3}} = z^{\frac{8}{3}}$
Independent Practice (11-20)
Step1: Power of a power rule
For $(y^2)^{39}$, multiply exponents: $2 \times 39 = 78$, so $y^{2 \times 39} = y^{78}$
Step2: Power of a power rule
For $(4^2)^3$, multiply exponents: $2 \times 3 = 6$, so $4^{2 \times 3} = 4^6$
Step3: Simplify inside parentheses first
Subtract exponents: $\frac{6^4}{6^2} = 6^{4-2} = 6^2$. Apply power of a power: $(6^2)^3 = 6^{2 \times 3} = 6^6$
Step4: Simplify inside root first
Subtract exponents: $\frac{x^4y^2}{x^2y} = x^{4-2}y^{2-1} = x^2y^1$. Apply square root: $(x^2y^1)^{\frac{1}{2}} = x^{2 \times \frac{1}{2}}y^{1 \times \frac{1}{2}} = x^1y^{\frac{1}{2}}$
Step5: Apply square root to each term
$\sqrt{s^2m^2h^2} = (s^2m^2h^2)^{\frac{1}{2}} = s^{2 \times \frac{1}{2}}m^{2 \times \frac{1}{2}}h^{2 \times \fra…
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Guided Practice:
- $4^6$
- $b^{56}$
- $4^1$
- $x^{\frac{3}{2}}$
- $2^{\frac{4}{3}}$
- $x^{\frac{5}{3}}$
- $s^{\frac{3}{3}}m^{\frac{6}{3}}h^{\frac{3}{3}} = s^1m^2h^1$
Independent Practice (1-10):
- $(-2)^9$
- $2^{-4}$
- $x^{-6}$
- $x^1$
- $a^1b^1c^1$
- $3^6$
- $x^{30}$
- $2^{\frac{4}{3}}$
- $x^{\frac{3}{2}}y^2z^2$
- $z^{\frac{8}{3}}$
Independent Practice (11-20):
- $y^{78}$
- $4^6$
- $6^6$
- $x^1y^{\frac{1}{2}}$
- $s^1m^1h^1$
- $2^2$
- $3^{\frac{7}{2}}$
- $x^{12}$
- $x^{\frac{5}{3}}y^{\frac{5}{3}}$
- $x^3$