Simplify. $\dfrac{a^3}{3a^2}\times 27a^6$ Answer ... $\sqrt{\dfrac{4}{16x^3}}$ Answer ... $\dfrac{49^{2a}}{7^b}=343^{4c}$. Find an expression for $b$ in terms of $a$ and $c$. Answer ...

1. Simplify the algebraic fraction first.

$\frac{a^3}{3a^2}=\frac{1}{3}a^{3-2}=\frac{a}{3}$

2. Multiply by $27a^6$ and simplify fully.

$\frac{a}{3}\times 27a^6=9a^{1+6}=9a^7$

1. Simplify the numerical fraction inside the square root.

$\sqrt{\dfrac{4}{16x^3}}=\sqrt{\dfrac{1}{4x^3}}$

2. Take the square root and simplify the surd form.

$\sqrt{\dfrac{1}{4x^3}}=\dfrac{1}{\sqrt{4x^3}}=\dfrac{1}{2x\sqrt{x}}=\dfrac{\sqrt{x}}{2x^2}$

1. Rewrite 49 and 343 as powers of 7.

Since $49=7^2$ and $343=7^3$,
$\frac{49^{2a}}{7^b}=343^{4c}$
becomes
$\frac{(7^2)^{2a}}{7^b}=(7^3)^{4c}.$

2. Simplify the indices on both sides using laws of indices.

$(7^2)^{2a}=7^{4a}, \qquad (7^3)^{4c}=7^{12c}$
So
$\frac{7^{4a}}{7^b}=7^{12c}.$
Using $\dfrac{7^m}{7^n}=7^{m-n}$,
$7^{4a-b}=7^{12c}.$

3. Equate the indices and solve for $b$.

Since the bases are equal,
$4a-b=12c.$
Hence
$-b=12c-4a$
$b=4a-12c.$