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
Solve
3^2x + 3^x - 21
check extraneous roots
4 answers
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.
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.
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.
1 answer