e^x-4e^-x=3 how would i solve this equation?

Question

campbell_st · Answer

so the equation is 

$$e^x - \frac{4}{e^x} = 3$$

multiply every term by $$e^x$$

then you get 

$$e^{2x} - 4 = 3e^x$$

or 

$$e^{2x} -3x^x - 4 = 0$$

now you have an equation that can be reduced to a quadratic... 

by letting 

$$e^x = u$$

so you have 

$$u^2 - 3u - 4 = 0$$

solve for u, then make the reverse substitution and then solve for x. 

hope it helps

anonymous · Answer

@campbell_st  thanks , so would the answer be u=4?

campbell_st · Answer

well you factor and have 

(u + 1)(u - 4) = 0

so u = 4 or -1

then the reverse substitution 

$$e^x = 4 ~~~and~~~e^x = -1$$

so you now need to decide on the solution, and you have made a sound decision based on what you have posted. 

so now solve for x

anonymous · Answer

So then would you just substitute 4 in ? 4-4/4=3?

anonymous · Answer

$$e^{-1}$$ has no solution, whereas if $e^x=4$ then $x=\ln(4)$