Solve the equation
-5e^-x + 9 = 6
I subtracted 9 to both sides and got:
-5e^-x= -3 < this is where I'm stuck at.
- 5e ^ ( - x ) + 9 = 6
- 5e ^ ( - x ) = 6 - 9
- 5e ^ ( - x ) = - 3 Divide both sides by - 5
e ^ ( - x ) = 3 / 5
Take the natural logarithm of both sides:
- x = - log (5 / 3 )
Divide both sides by -1:
x = log ( 5 / 3 )
x = 0.510826
Remark:
log mean natural logarithm
Tysm :D
To solve the equation -5e^-x + 9 = 6, you correctly subtracted 9 from both sides, which gives you -5e^-x = -3.
To continue solving for x, you need to isolate the variable e^-x. To do that, divide both sides of the equation by -5:
(-5e^-x) / -5 = -3 / -5
Simplifying the equation, you get:
e^-x = 3/5
Now, to solve for x, you need to take the natural logarithm (ln) of both sides. The natural logarithm is the inverse of the exponential function e^x. Applying the natural logarithm to both sides gives:
ln(e^-x) = ln(3/5)
Using the property that ln(e^a) = a, the equation simplifies to:
-x = ln(3/5)
Finally, to solve for x, multiply both sides of the equation by -1 to get:
x = -ln(3/5)
Therefore, the solution to the equation -5e^-x + 9 = 6 is x = -ln(3/5).