Solve
3^2x + 3^x - 21
check extraneous roots
Is the first term 3^2x (which is 9x) or 3 x^2 or 3^(2x)?
You have not stated an equation, only a function definition or "expression". There is nothing to "solve"
The solution to
3^(2x) + 3^x -21 = 0 is about x = 1.29
how was it solved ? quad formula. what is a b and c?
If you are talking about using a quadratic formula, you wrote the equation wrong. As i said before, you did not even write and equation
It should have been 3x^2 + 3x -21 = 0, whic is the same as x^2 +x -7 = 0
That equation has only irrational roots.
I solved the other equation by trial and error.
Here is a way to solve
3^(2x) + 3^x -21 = 0
Let 3^x = y
Then,
y^2 + y -21 = 0
b^2 -4ac = 85
The roots are:
y = [-1 +/- sqrt 85]/2
= -5.1098 or 4.1098
When y = 4.1098,
x = (lny)/ln3 = 1.2865
I mentioned that solution in a previous post.
There is another "y" solution corresponding to y = -5.1098
However that appears to require the log of a negative number, which has no real solution for x.
To solve the equation 3^(2x) + 3^x - 21, we can use a substitution to simplify it. Let's substitute y = 3^x. This means that y^2 = (3^x)^2 = 3^(2x).
Applying this substitution in the equation, we have:
y^2 + y - 21 = 0
Now, we can solve this quadratic equation for y using factoring or the quadratic formula. Factoring method is more convenient in this case, so let's factor it:
(y - 3)(y + 7) = 0
Setting each factor equal to zero, we have two possible values for y:
y - 3 = 0 -> y = 3
y + 7 = 0 -> y = -7
Since y = 3^x, we get two possible solutions for x:
3^x = 3 -> x = 1
3^x = -7 (No solution, as 3^x is always positive)
However, we need to check for any extraneous roots. In this case, it means verifying if x = 1 is a valid solution by substituting it back into the original equation and seeing if it holds true.
When x = 1,
3^(2x) + 3^x - 21 = 3^(2*1) + 3^1 - 21
= 3^2 + 3 - 21
= 9 + 3 - 21
= -9
Since -9 is not equal to 0, the solution x = 1 does not satisfy the original equation. Therefore, there are no valid solutions for this equation.