Given the equation e^(-4x) = 6.
We need to find x values where the equality holds.
To solve the exponent equation, will apply the natural logarithm to both sides:
==> ln e^(-4x) = ln 6
Now we know from logarithm properties that:
ln a^b = b*ln a
==> (-4x) * ln e = ln 6
But ln e = 1
==> -4x *1 = ln 6
==> -4x = ln 6
Now divide by -4:
==> x = ln 6 / -4
==> x = -0.4479
We'll recall the principle that:
e^a = b <=> a = ln b
For the given equation, we'll take logarithms both sides:
ln e^(-4x) = ln 6
We'll apply the power property of logarithms:
ln e^a = a*ln e
-4x*ln e = ln 6
But ln e = 1.
-4x = ln 6
We'll divide by -4 both sides:
x = -ln 6/4
We'll get the calculator to find ln 6 = 1.7917
x = -1.7917/4
The solution of x, rounded to four decimal places, is:
x = -0.4479
To find the solution to e^(-4x) = 6.
e^(-4x) = 6.
We take natural logarithms of both sides:
-4x = ln 6.
-4x = 1.791759 469.
We divide both sides by -4:
x = 1.791759 469/-4
x = -0.4479 for 4 decimal places.