Question

8
The number of integers satisfying the inequality log 1/2 |x-3|>-1 is
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Answer 1

We're given the inequality,

$\longrightarrow\log_{\frac{1}{2}}|x-3|>-1$

Since $\log x$ is only defined for $x>0,$

$\longrightarrow |x-3|>0$

On removing the modulus sign in LHS we get the following.

$\longrightarrow x-3\in(-\infty,\ 0)\cup(0,\ \infty)$

$\longrightarrow x\in(-\infty,\ 3)\cup(3,\ \infty)\quad\quad\dots(1)$

Now,

$\longrightarrow\log_{\frac{1}{2}}|x-3|>-1$

or,

$\longrightarrow\dfrac{\log|x-3|}{\log\left(\dfrac{1}{2}\right)}>-1$

$\longrightarrow\dfrac{\log|x-3|}{-\log2 }>-1\quad\quad\left[\because\,\log\left(\dfrac{1}{2}\right)=\log\left(2^{-1}\right)=-\log2\right]$

$\longrightarrow-\dfrac{\log|x-3|}{\log2} >-1$

Multiplying by -1, (note the symbol change)

$\longrightarrow\dfrac{\log|x-3|}{\log2}<1$

Multiplying by $\log2,$

$\longrightarrow\log|x-3|<\log2$

Taking antilog,

$\longrightarrow|x-3|<2$

On removing the modulus sign in LHS we get the following.

$\longrightarrow x-3\in(-2,\ 2)$

$\longrightarrow x\in(-2+3,\ 2+3)$

$\longrightarrow x\in(1,\ 5)\quad\quad\dots(2)$

Taking $(1)\land(2)$ we get,

$\longrightarrow x\in(1,\ 3)\cup(3,\ 5)$

This is the solution of the inequality.

So the possible integral values of $x$ are,

$\longrightarrow x\in\{2,\ 4\}$

Hence the no. of integral values of $x$ is 2.

Answer 2

hello

Answer

log

2

1

∣x−3∣ is meaningful when x



=3

log

2

1

∣x−3∣>−1

⇒−log

2

∣x−3∣>−1[∵log

1/a

b=−log

a

b]

⇒log

2

∣x−3∣<1, multiplied both sides with −1, so inequality changes

⇒∣x−3∣<2

⇒−2<x−3<2

⇒1<x<5

⇒x∈(1,5)−{3}

Therefore, integral values are 2,4 that is 2 integral values.

Ans: 2