Solve the equation. Check for extraneous roots.
4^2x-4^x-20=0
Notice that we can write 4^(2x) as (4^x)^2
and you equation becomes a quadratic
let 4^x = k
then your equation becomes:
k^2 - k - 20 = 0
(k-5)(k+4) = 0
k = 5 or k = -4
then 4^x = 5
log 4^x = log 5
x log4 = log 5
x = log5/log4 = appr 1.161
and 4^x = -4
x log4 = log (-4), but log -4 would be undefined , so this part leads to an extraneous root.
x = appr 1.161
To solve the equation 4^(2x) - 4^x - 20 = 0, we can use a substitution. Let y = 4^x. Then the equation becomes:
y^2 - y - 20 = 0.
This is a quadratic equation that we can factor or solve using the quadratic formula. Factoring, we have:
(y - 5)(y + 4) = 0.
Setting each factor equal to zero gives us two possible values for y:
y - 5 = 0 --> y = 5,
y + 4 = 0 --> y = -4.
Now we substitute back to find the values of x:
For y = 5, we have:
4^x = 5,
x = log base 4 (5).
For y = -4, we have:
4^x = -4,
This equation does not have a real solution since raising any number to a positive power will always yield a positive result.
So the only solution is x = log base 4 (5).
To check for extraneous roots, we substitute x = log base 4 (5) back into the original equation:
4^(2 * log base 4 (5)) - 4^(log base 4 (5)) - 20 = 0.
Using exponentiation rules, this simplifies to:
(4^log base 4 (5))^2 - 5 - 20 = 0,
5^2 - 5 - 20 = 0,
25 - 5 - 20 = 0,
0 = 0.
This confirms that x = log base 4 (5) is indeed a valid solution and there are no extraneous roots in this case.
To solve the equation 4^2x - 4^x - 20 = 0, we can create a substitution to simplify the equation. Let's substitute a term to make the equation more manageable.
Let's say we substitute 4^x with a single variable, such as "y." Then, we can rewrite the equation as y^2 - y - 20 = 0.
Now, we have a quadratic equation (a standard form of ax^2 + bx + c = 0) that we can solve by factoring, completing the square, or using the quadratic formula.
In this case, we can easily factor the quadratic equation, y^2 - y - 20 = 0, by splitting the middle term.
The factors of -20 that add up to -1 (the coefficient of the linear term) are -5 and 4. So, we can rewrite the equation as (y - 5)(y + 4) = 0.
Now, we can set each factor equal to zero and solve for y:
y - 5 = 0 ---> y = 5
y + 4 = 0 ---> y = -4
Thus, we have two possible values for y: 5 and -4.
Remember that we substituted y = 4^x.
Substituting both values in, we get:
For y = 5: 4^x = 5, which gives x ≈ 0.8614 (using logarithms).
For y = -4: 4^x = -4, which is not possible since 4^x is always positive for any real value of x.
Now, we need to check for extraneous roots, which are values that satisfy the equation but are not valid solutions. In this case, 4^x cannot be negative or zero.
Therefore, the only valid solution is x ≈ 0.8614.