# If (1/125)^(a^2 + 4ab) = (3sqrt625)^(3a^2 - 10ab) and if a and b do not equal 0.. a/b = ........ ?

Aug 22, 2017

See below.

#### Explanation:

Applying $\log$ to both sides

$\left({a}^{2} + 4 a b\right) L o {g}_{e} \left(\frac{1}{125}\right) - \left(3 {a}^{2} - 10 a b\right) L o {g}_{e} \left(3 \times 25\right) = 0$

Now solving for $a$ we get at

$\left\{\begin{matrix}a = 0 \\ a = \frac{2 b L o {g}_{e} \left({3}^{5} {5}^{4}\right)}{3 L o {g}_{e} \left(3 \times {5}^{3}\right)}\end{matrix}\right.$

and finally

$\frac{a}{b} = \frac{2 {\log}_{e} \left({3}^{5} {5}^{4}\right)}{3 {\log}_{e} \left(3 \times {5}^{3}\right)} \approx 1.34199$

Aug 24, 2017

$\frac{a}{b} = \frac{4}{21}$

#### Explanation:

Using Indices

${\left(\frac{1}{125}\right)}^{{a}^{2} + 4 a b} = {\left(\sqrt[3]{625}\right)}^{3 {a}^{2} - 10 a b}$

Recall that $\Rightarrow \frac{1}{a} = {a}^{-} 1$

${\left({125}^{-} 1\right)}^{{a}^{2} + 4 a b} = {\left({625}^{\frac{1}{3}}\right)}^{3 {a}^{2} - 10 a b}$

${\left({5}^{3 \left(- 1\right)}\right)}^{{a}^{2} + 4 a b} = {\left({5}^{4 \left(\frac{1}{3}\right)}\right)}^{3 {a}^{2} - 10 a b}$

${5}^{- 3 \left({a}^{2} + 4 a b\right)} = {5}^{\frac{4}{3} \left(3 {a}^{2} - 10 a b\right)}$

${\cancel{5}}^{- 3 \left({a}^{2} + 4 a b\right)} = {\cancel{5}}^{\frac{4}{3} \left(3 {a}^{2} - 10 a b\right)}$

$- 3 \left({a}^{2} + 4 a b\right) = \frac{4}{3} \left(3 {a}^{2} - 10 a b\right)$

$- 3 {a}^{2} - 12 a b = \frac{12 {a}^{2}}{3} - \frac{40 a b}{3}$

$- 3 {a}^{2} - 12 a b = \frac{12 {a}^{2} - 40 a b}{3}$

Cross Multiply

$3 \left(- 3 {a}^{2} - 12 a b\right) = 12 {a}^{2} - 40 a b$

$- 9 {a}^{2} - 36 a b = 12 {a}^{2} - 40 a b$

Collect Like Terms

$- 9 {a}^{2} - 12 {a}^{2} = - 40 a b + 36 a b$

$- 21 {a}^{2} = - 4 a b$

$\cancel{-} 21 {a}^{2} = \cancel{-} 4 a b$

$21 {a}^{2} = 4 a b$

Since we are looking for $\textcolor{w h i t e}{x} \frac{a}{b}$

Divide both sides by $a b$

$\frac{21 {\cancel{{a}^{2}}}^{1}}{\cancel{a} b} = \frac{4 \cancel{a b}}{\cancel{a b}}$

$\Rightarrow \frac{21 a}{b} = 4$

Divide both sides by $21$

$\Rightarrow \frac{\frac{21 a}{b}}{21} = \frac{4}{21}$

$\Rightarrow \left(\frac{\cancel{21} a}{b}\right) \times \frac{1}{\cancel{21}} = \frac{4}{21}$

$\Rightarrow \frac{a}{b} = \frac{4}{21} \to A n s w e r$

Hence Option. A is the final Answer..