## anonymous 5 years ago how do you solve 4+5e^-x=0

1. Owlfred

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2. anonymous

Rearrange and take logarithm on both sides! :)

3. anonymous

i still don understand

4. anonymous

Subtracting 4 from both sides gives $$5e^{-x}=-4$$. Now divide both sides by 5, you get $$e^{-x}=- \frac{4}{5}$$. But the exponent function has only positive values in $$\ R$$. So, the equation has no real solutions.

5. anonymous

exponential function*

6. anonymous

u understood amberllg?