How do you solve $5^{- x} = 3$ $5^-x = 3$ ?

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Answer 1 · 2016-06-03T19:31:44

Using the log.

To solve the equations where the $x$ $x$ is in the exponent, the trick is to do the logarithms.

$5^{- x} = 3$ $5^(-x)=3$

$log (5^{- x}) = log (3)$ $log(5^(-x))=log(3)$

we then apply the property of the logarithm where the exponent in the log goes in front multiplying.

$- x log (5) = log (3)$ $-xlog(5)=log(3)$

$x = - \frac{log (3)}{log (5)} \approx - 0.68$ $x=-log(3)/log(5)\approx-0.68$ .

How do you solve 5−x=35^-x = 3?