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CAT 2024 Slot 3 — QA Question 4

Solving Quadratic EquationsEasy

The sum of all distinct real values of x that satisfy the equation 10x + 410x = 812, is

Answer & solution

Correct answer: 2log 10 2

  • A

    3log102

  • 2log102

  • C

    4log102

  • D

    log102

Solution

Medium

The equation 10x+410x=81210^{x}+\dfrac{4}{10^{x}}=\dfrac{81}{2} is a quadratic in t=10xt=10^{x}. The two solutions x1,x2x_1,x_2 correspond to the two roots t1,t2t_1,t_2; their sum is log⁡10(t1t2)\log_{10}(t_1t_2), and the product of roots comes straight from the quadratic.

1

Substitute and form a quadratic. Let t=10x>0t=10^{x}>0.

t+4t=812⇒ 2t2−81t+8=0(multiply by 2t)\begin{aligned} &t+\frac{4}{t}=\frac{81}{2}\\ &\Rightarrow\ 2t^2-81t+8=0\quad\text{(multiply by }2t\text{)} \end{aligned}
2

Product of roots. For 2t2−81t+8=02t^2-81t+8=0, the product of roots is 82=4\tfrac{8}{2}=4 (both roots positive, so both give valid xx).

t1t2=82=4(product =c/a)\begin{aligned} &t_1 t_2=\frac{8}{2}=4\quad\text{(product }=c/a\text{)} \end{aligned}
3

Sum of the xx-values. Since x=log⁡10tx=\log_{10}t, adding the two solutions converts the product into a sum of logs.

x1+x2=log⁡10t1+log⁡10t2⇒ x1+x2=log⁡10(t1t2)(log of a product)⇒ x1+x2=log⁡104(from step 2)⇒ x1+x2=2log⁡102\begin{aligned} &x_1+x_2=\log_{10}t_1+\log_{10}t_2\\ &\Rightarrow\ x_1+x_2=\log_{10}(t_1 t_2)\quad\text{(log of a product)}\\ &\Rightarrow\ x_1+x_2=\log_{10}4\quad\text{(from step 2)}\\ &\Rightarrow\ x_1+x_2=2\log_{10}2 \end{aligned}
2log⁡1022\log_{10}2

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CAT 2024 Slot 3 QA Q4 — Solving Quadratic Equations solution | TheCATExam