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11: Solve the following exponential equation\newline4x+2x+212=04^{x}+2^{x+2}-12=0

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Q. 11: Solve the following exponential equation\newline4x+2x+212=04^{x}+2^{x+2}-12=0
  1. Identify Equation Structure: Let's first identify the structure of the equation and see if we can simplify it using properties of exponents.\newlineThe equation is 4x+2x+212=04^{x} + 2^{x+2} - 12 = 0.\newlineWe notice that 44 is a power of 22, specifically 4=224 = 2^2, so we can rewrite 4x4^{x} as (22)x=22x(2^2)^{x} = 2^{2x}.\newlineAlso, 2x+22^{x+2} can be written as 2x22=42x2^x \cdot 2^2 = 4 \cdot 2^x.\newlineNow, let's rewrite the equation with these observations.
  2. Rewrite Using Exponents: Rewrite the equation using the properties of exponents. 4x+2x+212=04^{x} + 2^{x+2} - 12 = 0 becomes 22x+42x12=02^{2x} + 4 \cdot 2^x - 12 = 0. Now, we have an equation that resembles a quadratic equation in form of 2x2^x. Let's set y=2xy = 2^x and rewrite the equation in terms of yy.
  3. Substitute and Rewrite: Substitute y=2xy = 2^x into the equation.\newlineThe equation becomes y2+4y12=0y^2 + 4y - 12 = 0.\newlineNow, we have a quadratic equation in yy, which we can solve using factoring, completing the square, or the quadratic formula.\newlineLet's try to factor the quadratic equation.
  4. Factor Quadratic Equation: Factor the quadratic equation y2+4y12=0y^2 + 4y - 12 = 0. We look for two numbers that multiply to 12-12 and add up to 44. The numbers 66 and 2-2 satisfy these conditions. So, we can write the equation as (y+6)(y2)=0(y + 6)(y - 2) = 0. Now, we can find the values of yy by setting each factor equal to zero.
  5. Solve for y: Solve for y from the factors (y+6)=0(y + 6) = 0 and (y2)=0(y - 2) = 0. For the first factor, y+6=0y + 6 = 0, we get y=6y = -6. For the second factor, y2=0y - 2 = 0, we get y=2y = 2. Now we have two possible values for y, which are y=6y = -6 and y=2y = 2.
  6. Solve for x: Recall that y=2xy = 2^x. We will now solve for xx using the values of yy we found.\newlineFirst, let's take y=6y = -6. Since 2x2^x is always positive, there is no real number xx such that 2x=62^x = -6. Therefore, y=6y = -6 does not give us a valid solution for xx.\newlineNow, let's take y=2y = 2. We need to solve xx00.
  7. Solve for xx: Recall that y=2xy = 2^x. We will now solve for xx using the values of yy we found.\newlineFirst, let's take y=6y = -6. Since 2x2^x is always positive, there is no real number xx such that 2x=62^x = -6. Therefore, y=6y = -6 does not give us a valid solution for xx.\newlineNow, let's take y=2xy = 2^x00. We need to solve y=2xy = 2^x11.Solve the equation y=2xy = 2^x11.\newlineSince the bases are the same, we can equate the exponents.\newlineThis gives us y=2xy = 2^x33.\newlineSo, the solution to the equation y=2xy = 2^x11 is y=2xy = 2^x33.

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