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CAT 2019 Slot 2 — QA Question 1

Solving Quadratic EquationsEasy

The real root of the equation 26x + 23x+2 - 21 = 0 is

Answer & solution

Correct answer: log 2 3 3

  • log2(3)3

  • B

    log29

  • C

    log2(7)3

  • D

    log227

Solution

Easy

The exponents are multiples of 3x3x, so substitute y=23xy=2^{3x} to turn the equation into a quadratic in yy. Solve it, discard the impossible (negative) root, then read off xx with a logarithm.

1

Substitute to get a quadratic. Let y=23xy=2^{3x}, so 26x=y22^{6x}=y^{2} and 23x+2=22⋅23x=4y2^{3x+2}=2^{2}\cdot 2^{3x}=4y.

26x+23x+2−21=0⇒ y2+4y−21=0(with y=23x)\begin{aligned} &2^{6x}+2^{3x+2}-21=0\\ &\Rightarrow\ y^{2}+4y-21=0 \quad\text{(with }y=2^{3x}\text{)} \end{aligned}
2

Solve the quadratic. Factor the equation from step 1.

y2+4y−21=0⇒ (y+7)(y−3)=0⇒ y=−7  or  y=3\begin{aligned} &y^{2}+4y-21=0\\ &\Rightarrow\ (y+7)(y-3)=0\\ &\Rightarrow\ y=-7 \ \ \text{or}\ \ y=3 \end{aligned}
3

Reject the negative root. Since y=23xy=2^{3x} is a power of 22, it is always positive, so y=−7y=-7 is impossible.

23x=3(only valid root)\begin{aligned} &2^{3x}=3 \quad\text{(only valid root)} \end{aligned}
4

Take logarithms (base 2). Solve 23x=32^{3x}=3 from step 3 for xx.

3x=log⁡23⇒ x=log⁡233\begin{aligned} &3x=\log_{2}3\\ &\Rightarrow\ x=\frac{\log_{2}3}{3} \end{aligned}
x=log⁡233x=\dfrac{\log_{2}3}{3}

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CAT 2019 Slot 2 QA Q1: The real root of the equation 2 6x + 2 3x+2 - 21 = 0 is — Solution | TheCATExam