Solve for x:
1. 3^2x+3^x+1-4=0
2. log3(x=12)+log3(x+4)=2
3. lnx+ln(x+2)=4
4. 3*4^x+4*2^x+8=0
1. To solve for x in the equation 3^(2x) + 3^x + 1 - 4 = 0, we can use a substitution method. Let's substitute a variable, say u, to simplify the equation.
Let u = 3^x. Rearranging the equation, we have u^2 + u - 3 = 0.
Now, we can solve this quadratic equation for u. Using factoring or the quadratic formula, we find that u = 1 or u = -3.
Substituting back u = 3^x, we have two equations:
1) 3^x = 1
Taking the logarithm of both sides, we get x = log3(1) = 0.
2) 3^x = -3
Since a positive number raised to any power will always be positive, there is no solution for this equation.
Therefore, the solution for x is x = 0.
2. To solve for x in the equation log3(x + 12) + log3(x + 4) = 2, we can use the logarithmic properties.
First, we apply the product rule of logarithms:
log3((x + 12)(x + 4)) = 2.
Next, we can rewrite the equation in exponential form:
3^2 = (x + 12)(x + 4).
Simplifying, we have:
9 = x^2 + 16x + 48.
Rearranging the equation and setting it equal to zero:
x^2 + 16x + 39 = 0.
Now we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. After solving, we find that x = -13 or x = -3.
However, we need to check these solutions. Plugging x = -13 and x = -3 back into the original equation, we see that both solutions are extraneous, resulting in a negative logarithm.
Therefore, this equation has no real solutions.
3. To solve for x in the equation ln(x) + ln(x + 2) = 4, we can combine the logarithms using the properties of logarithms.
Using the logarithmic property ln(a) + ln(b) = ln(a * b), the equation simplifies to:
ln(x(x + 2)) = 4.
Rewriting in exponential form, we have:
x(x + 2) = e^4.
Now, we can solve the quadratic equation. Expanding and rearranging, we have:
x^2 + 2x = e^4.
Rearranging the equation and setting it to zero:
x^2 + 2x - e^4 = 0.
To solve this quadratic equation, you can use factoring, completing the square, or the quadratic formula. After solving, you will find the values of x.
Please note that depending on the value of e^4, the equation may have two real solutions, one real solution, or no real solutions.
4. To solve for x in the equation 3 * 4^x + 4 * 2^x + 8 = 0, we can use factoring methods.
The equation cannot be simplified by combining terms or applying logarithmic properties, so we proceed with factoring.
Rearranging the equation, we have:
3(4^x) + 4(2^x) + 8 = 0.
Applying the distributive property:
4(2^x)(2^x) + 4(2^x) + 8 = 0.
Substituting y = 2^x for simplicity, we have:
4(y^2) + 4y + 8 = 0.
Dividing each term by 4:
y^2 + y + 2 = 0.
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. After solving, we find the values of y.
Substituting y back as 2^x:
2^x = (solutions for y).
Finally, solve for x by taking the logarithm base 2 of both sides:
x = log2(2^x).
Remember to check the solutions and their validity within the original equation, as sometimes extraneous solutions may emerge.