3^2x - 11(3^x) + 30 = 0 ; solve for x.
I know that you can substitute 3^x with another variable like y to make it easier, so it looks like this:
y^2 - 11y + 30 = 0
...unless I did something wrong there. Help please?
Thanks!
great so far
now
y^2 - 11y + 30 = 0
(y-5)(y-6) = 0
so y = 5 or y = 6
but y = 3^x
so
3^x = 5 or 3^x = 6
I will finish the first one
take ln of both sides for
3^x = 5
ln(3^x) = ln5
x(ln3) = ln5
x = ln5/ln3
now you try the second one.
You're on the right track! Substituting 3^x with another variable like y is a great strategy. Let's go through the steps to solve the equation:
1. Substitute 3^x with y:
y^2 - 11y + 30 = 0
2. Factor the quadratic equation:
(y - 6)(y - 5) = 0
3. Set each factor equal to zero and solve for y:
y - 6 = 0 or y - 5 = 0
Solving these equations gives us two potential solutions for y:
y = 6 or y = 5
4. Now, we need to substitute y back to 3^x:
If y = 6, then 3^x = 6
If y = 5, then 3^x = 5
5. Solve the equation 3^x = 6:
Taking the logarithm of both sides with base 3, we get:
xlog3(3) = log3(6)
x = log3(6)
6. Solve the equation 3^x = 5:
Taking the logarithm of both sides with base 3, we get:
xlog3(3) = log3(5)
x = log3(5)
So the solutions for x are:
x = log3(6)
x = log3(5)
These are the exact solutions for the given equation.