Question

7 mark for review if $x^{y}x^{6}=x^{54}$ and $(x^{3})^{z}=x^{9}$, what is the value of $y + z$? a 11 b 12 c 48 d 51 8 mark for review which of the following expressions is equivalent to $sqrt4{81b^{3}c}$? a $3b^{\frac{3}{4}}c^{\frac{1}{4}}$ b $3b^{3}c$ c $20.25b^{\frac{3}{4}}c^{\frac{1}{4}}$ d $20.25b^{3}c$ 9 mark for review if $x^{\frac{3}{2}} = 8x$, which of the following could be the value of $x$? a 2 b 4 c 6 d 8 10 mark for review which of the following expressions is equivalent to $(3m^{2}n^{-3})^{\frac{2}{3}}$? a $2m^{\frac{4}{3}}n^{-2}$ b $sqrt3{9}m^{\frac{4}{3}}n^{-2}$ c $3m^{\frac{4}{3}}n^{-2}$ d $3m^{4}n^{-6}$

Explanation:

7.

Step1: Use exponent - rule for $x^y x^6=x^{54}$

According to the rule $a^m\times a^n=a^{m + n}$, we have $x^{y + 6}=x^{54}$. So, $y+6 = 54$, and $y=54 - 6=48$.

Step2: Use exponent - rule for $(x^3)^z=x^9$

According to the rule $(a^m)^n=a^{mn}$, we have $x^{3z}=x^9$. So, $3z = 9$, and $z = 3$.

Step3: Calculate $y + z$

$y+z=48 + 3=51$.

Step1: Rewrite $\sqrt[4]{81b^3c}$

We know that $\sqrt[4]{81}= \sqrt[4]{3^4}=3$, and $\sqrt[4]{b^3c}=b^{\frac{3}{4}}c^{\frac{1}{4}}$. So, $\sqrt[4]{81b^3c}=3b^{\frac{3}{4}}c^{\frac{1}{4}}$.

Step1: Rearrange the equation $x^{\frac{3}{2}}=8x$

Move all terms to one - side: $x^{\frac{3}{2}}-8x = 0$. Factor out $x$: $x(x^{\frac{1}{2}} - 8)=0$.

Step2: Solve for $x$

Set each factor equal to zero. First, $x = 0$. Second, for $x^{\frac{1}{2}}-8 = 0$, we have $x^{\frac{1}{2}}=8$, then $x = 64$. But if we check in the original equation, when $x = 0$, the left - hand side is $0$ and the right - hand side is $0$. Also, if we rewrite the original equation as $\sqrt{x^3}=8x$. Squaring both sides (note that we need to check for extraneous solutions later) gives $x^3 = 64x^2$, $x^3-64x^2=0$, $x^2(x - 64)=0$. Another way is to rewrite $x^{\frac{3}{2}}=8x$ as $x^{\frac{3}{2}}-8x=0$, $x(x^{\frac{1}{2}} - 8)=0$. If $x
eq0$, then $x^{\frac{1}{2}}=8$, $x = 64$. If we assume non - zero $x$ and divide the original equation $x^{\frac{3}{2}}=8x$ by $x$ (since $x = 0$ is a solution we can consider separately), we get $x^{\frac{1}{2}}=8$, $x = 64$. But if we consider the domain and the original form, when $x = 0$ the equation holds. However, if we rewrite it in a non - zero dividing way, from $x^{\frac{3}{2}}=8x$ to $x^{\frac{1}{2}}=8$ (for $x
eq0$), we get $x = 64$. If we go back to the original and check values from the options, when $x = 64$, $64^{\frac{3}{2}}=\sqrt{64^3}=8^3 = 512$ and $8\times64 = 512$.

Answer:

D. 51

8.