How do you solve $- 5 e^{- x} + 9 = 6$ $-5e^(-x)+9=6$ ?

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Answer 1 · 2016-09-06T15:46:30

$x = - ln (\frac{3}{5})$ $x = - ln((3) / (5))$

We have: $- 5 e^{- x} + 9 = 6$ $- 5 e^(- x) + 9 = 6$

Let's begin by subtracting $9$ $9$ from both sides of the equation:

$\Rightarrow - 5 e^{- x} = - 3$ $=> - 5 e^(- x) = - 3$

Then, let's divide both sides by $- 5$ $- 5$ :

$\Rightarrow e^{- x} = \frac{3}{5}$ $=> e^(- x) = (3) / (5)$

Now, let's apply the natural logarithm to both sides:

$\Rightarrow ln (e^{- x}) = ln (\frac{3}{5})$ $=> ln(e^(- x)) = ln((3) / (5))$

Using the laws of logarithms:

$\Rightarrow - x ln (e) = ln (\frac{3}{5})$ $=> - x ln(e) = ln((3) / (5))$

$\Rightarrow - x \cdot 1 = ln (\frac{3}{5})$ $=> - x cdot 1 = ln((3) / (5))$

$\Rightarrow - x = ln (\frac{3}{5})$ $=> - x = ln((3) / (5))$

Finally, to solve for $x$ $x$ , let's divide both sides by $- 1$ $- 1$ :

$\Rightarrow x = - ln (\frac{3}{5})$ $=> x = - ln((3) / (5))$

How do you solve −5e−x+9=6-5e^(-x)+9=6?