How do you solve $7^{x} = 5^{x - 4}$ $7^x=5^(x-4)$ ?

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Answer 1 · 2015-08-22T08:20:53

Take $log$ $log$ of both sides and use properties of $log$ $log$ of exponents to find:

$x = \frac{- 4 log (5)}{log (7) - log (5)} \approx - 19.133$ $x = (-4log(5))/(log(7) - log(5)) ~~ -19.133$

Take $log$ $log$ of both sides to get:

$x log (7) = log (7^{x}) = log (5^{x - 4}) = (x - 4) log (5)$ $x log(7) = log(7^x) = log(5^(x-4)) = (x - 4) log(5)$

$= x log (5) - 4 log (5)$ $= x log(5) - 4 log(5)$

Subtract $x log (5)$ $x log(5)$ from both sides to get:

$x (log (7) - log (5)) = - 4 log (5)$ $x (log(7) - log(5)) = -4log(5)$

Divide both sides by $(log (7) - log (5))$ $(log(7) - log(5))$ to get:

$x = \frac{- 4 log (5)}{log (7) - log (5)} \approx - 19.133$ $x = (-4log(5))/(log(7) - log(5)) ~~ -19.133$

How do you solve 7^x=5^(x-4)7x=5x−47^x=5^(x-4)?