How do you solve $13^{- 9 x} = 16^{x - 2}$ $13^ { - 9x } = 16^ { x - 2}$ ?

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Answer 1 · 2017-11-27T19:47:56

$x = 0.214$ $x=0.214$

Take the log of both sides and solve as you would a normal equation.

$13^{- 9 x}$ $13^{-9x$ = $16^{x - 2}$ $16^{x-2}$

ln $13^{- 9 x}$ $13^{-9x}$ =ln $16^{x - 2}$ $16^{x-2}$

$- 9 x$ $-9x$ ln $(13) = (x - 2)$ $(13)=(x-2)$ ln16

$- 9 x \cdot 2.56 = (x - 2) \cdot 2.77$ $-9x * 2.56 = (x-2) * 2.77$

$- 23.08 x = 2.77 x - 5.54$ $-23.08x = 2.77x - 5.54$

$- 25.85 x = - 5.54$ $-25.85x = -5.54$

$x = 0.214$ $x = 0.214$ (approx when using 2 decimal places)

How do you solve 13−9x=16x−213^ { - 9x } = 16^ { x - 2}?