Question

If ${\left(\frac{1}{125}\right)}^{a^2+4ab}=(\sqrt[3]{3625}{)}^{3a^2-10ab}$ and if $a$ and $b$ do not equal $0$, then $\frac{a}{b}=$?

• A. $\frac{4}{21}$
• B. $2$
• C. $\frac{76}{21}$
• D. $4$
• E. $\frac{76}{3}$

Answer 1

The correct option is A $\frac{4}{21}$

Rewrite the given equation with the common base as follows:

${\left(\frac{1}{125}\right)}^{a^2+4ab}={\left(\sqrt[3]{3625}\right)}^{3a^2-10ab}\Rightarrow {\left(\frac{1}{(5{)}^3}\right)}^{a^2+4ab}={\left(\sqrt[3]{3(5{)}^4}\right)}^{3a^2-10ab}\Rightarrow {\left(\right(5{)}^{-3})}^{a^2+4ab}={\left(\right(5{)}^{\frac{4}{3}})}^{3a^2-10ab}\Rightarrow {\left(5\right)}^{-3(a^2+4ab)}={\left(5\right)}^{\frac{4}{3}(3a^2-10ab)}$

Setting the exponents equal as shown below:

$-3(a^2+4ab)=\frac{4}{3}(3a^2-10ab)\Rightarrow -3a^2-12ab=\frac{4}{3}(3a^2-10ab)\Rightarrow 3(-3a^2-12ab)=4(3a^2-10ab)\Rightarrow -9a^2-36ab=12a^2-40ab\Rightarrow 12a^2+9a^2=40ab-36ab\Rightarrow 21a^2=4ab\Rightarrow 21a=4b\Rightarrow \frac{a}{b}=\frac{4}{21}$