Solve for x: 9^x-3^x-8=0
To solve the equation 9^x - 3^x - 8 = 0, we can use a substitution to simplify it. Let's substitute a new variable and make the equation easier to solve.
Let's substitute y = 3^x. By doing this, we have:
9^x - y - 8 = 0.
Now, we can rewrite this equation as:
y^2 - y - 8 = 0.
We can solve this quadratic equation by factoring or by using the quadratic formula. Factoring may not be possible here, so let's use the quadratic formula:
The quadratic formula is given by:
y = (-b ±sqrt(b^2 - 4ac))/(2a), where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c = 0.
For our equation, a = 1, b = -1, and c = -8:
y = (-(-1) ± sqrt((-1)^2 - 4(1)(-8)))/(2(1)).
Simplifying:
y = (1 ± sqrt(1 + 32))/2.
Continuing:
y = (1 ± sqrt(33))/2.
Now, let's substitute back y = 3^x:
3^x = (1 ± sqrt(33))/2.
To isolate x, we take the logarithm of both sides. Let's use the natural logarithm, ln:
ln(3^x) = ln((1 ± sqrt(33))/2).
Using logarithm properties, we can bring down the exponent:
x ln(3) = ln((1 ± sqrt(33))/2).
Finally, solving for x:
x = ln((1 ± sqrt(33))/2) / ln(3).
Therefore, x = ln((1 + sqrt(33))/2) / ln(3) or x = ln((1 - sqrt(33))/2) / ln(3).
To solve the equation 9^x - 3^x - 8 = 0, we can observe that both 9 and 3 are powers of 3 (9 = 3^2 and 3 = 3^1). So, let's substitute 3^x with a variable, let's say t. Now the equation becomes:
t^2 - t - 8 = 0
To solve this quadratic equation, we can factorize it or use the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Where a = 1, b = -1, and c = -8. Substituting these values, we have:
t = (1 ± √((-1)^2 - 4 * 1 * -8)) / 2 * 1
t = (1 ± √(1 + 32)) / 2
t = (1 ± √33) / 2
Now, we substitute t back to 3^x:
3^x = (1 ± √33) / 2
To solve for x, we can take the logarithm log base 3 of both sides:
log3(3^x) = log3((1 ± √33) / 2)
Using the logarithmic property logb(b^a) = a, we get:
x = log3((1 ± √33) / 2)
Therefore, the solution to the equation 9^x - 3^x - 8 = 0 is:
x = log3((1 ± √33) / 2)