# Thread: Algebra Problem with Exponents

1. ## Algebra Problem with Exponents

$\dfrac{3}{4} t^{-1/4} = \dfrac{1}{4} t^{-1/3}$ hint? How to solve for t?

2. ## Re: Algebra Problem with Exponents

cross multiply and use correctly the rules for the indices...

3. ## Re: Algebra Problem with Exponents

Hello, Jason76!

$\text{Solve for }t\!:\;\tfrac{3}{4}\,\!t^{-1/4} = \tfrac{1}{4}\,\! t^{-1/3}$

Multiply by $4t^{\frac{1}{3}}\!:\;4t^{\frac{1}{3}}\left(\tfrac{ 3}{4}\,\!t^{-\frac{1}{4}}\right) \;= \;4t^{\frac{1}{3}}\left(\tfrac{1}{4}t^{-\frac{1}{3}}\right)$

. . . . . . . . . . . . . . . . . $3t^{\frac{1}{12}} \;=\;1$

. . . . . . . . . . . . . . . . . . $t^{\frac{1}{12}} \;=\;\frac{1}{3}$

. . . . . . . . . . . . . . . . . . . $t \;=\;\frac{1}{3^{^{12}}}$