Q128 / 200 · Manipal (MET) 2013 · Mathematics

$\log _2\left(9-2^x\right)=10^{\log (3-x)}$, solve for $x$

  • A
    3
  • B
    both (a) and (b)
  • C
    0 and 6
  • Correct answer
    D
    0

Solution

\(\begin{aligned} & \text {Given, } \log _2\left(9-2^x\right)=10^{\log _{10}(3-x)} \\ & \Rightarrow \log _2\left(9-2^x\right)=(3-x) \ldots\left[\because a^{\log _a b}=1\right] \\ & \Rightarrow 2^{3-x}=9-2^x \\ & \Rightarrow \frac{2^3}{2^x}=9-2^x \\ & \Rightarrow 8=2^x \times\left(9-2^x\right) \\ & \Rightarrow 2^{2 x}-2^x \times 9+8=0 \end{aligned}\) Let \(2^x=y\), then \(\begin{aligned} & y^2-9 y+8=0 \\ & \Rightarrow(y-8)(y-1)=0 \\ & \Rightarrow y=8 \text { or } y=1 \\ & \Rightarrow 2^x=2^3 \text { or } 2^x=2^{\circ} \\ & \Rightarrow x=3 \text { or } x=0 \end{aligned}\) But \(\mathrm{x}=3\) does not satisfy the given equation, since \(\log 0\) is not defined.

Hence, the correct answer is option: 'D'.

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