solve the equations for x
a) e^2x - e^x - 30 = 0
b)1-3/n^x = 5
let z = e^x
then
z^2 - z - 30 = 0
(z-6)(z+5) = 0
so
z = 6 or -5
if z = e^x = 6
x = ln 6 look it up
if z = -5
x = ln -5 impossible so only ln 6 is solution
Are you sure this is not a typo?
1-3/n^x = 5
-3/n^x = 4
n^x = -3/4
ln n^x = -ln .75
x ln n = -ln.75
x = -ln.75/ln n
x = .2877/ln n
a) To solve the equation e^2x - e^x - 30 = 0 for x, we can use a substitution. Let's substitute e^x with another variable, say a. So, the equation becomes a^2 - a - 30 = 0. Now we can solve this quadratic equation for a.
To factor the quadratic equation, we need to find two numbers that multiply to give -30 and add up to -1. The numbers -6 and 5 satisfy these conditions. So, we can rewrite the equation as (a - 6)(a + 5) = 0.
Setting each factor equal to zero, we have two possible solutions:
a - 6 = 0 -> a = 6
a + 5 = 0 -> a = -5
Now, we can substitute back e^x for a and solve for x:
e^x = 6 -> x = ln(6)
e^x = -5 -> No solution since e^x is always positive
Therefore, the solution to the equation e^2x - e^x - 30 = 0 is x = ln(6).
b) To solve the equation 1 - 3/n^x = 5 for x, we can start by isolating the term containing n^x.
Subtracting 1 from both sides, we get -3/n^x = 4.
Next, we can multiply both sides by -1 and flip the sides to get n^x/3 = -4.
Now, we will use logarithms to solve for x. Taking the logarithm of both sides (base n), we have log_n(n^x/3) = log_n(-4).
Using the logarithmic identity log_b(a^c) = c * log_b(a), the equation simplifies to x * log_n(n) - log_n(3) = log_n(-4).
Since log_n(n) = 1, the equation becomes x - log_n(3) = log_n(-4).
However, the logarithm of a negative number is undefined, so there is no solution to this equation.