Question

Q1.Number of real solution of the equation 
$\sqrt{log_{10}(-x)} = log_{10}\sqrt{x^{2} }$
​ is
(a) none
(b) exactly 1
(c) exactly 2
(d) 4



Nospams✍️​

Answer 1

Given,

$\small\text{ \longrightarrow\sqrt{\log_{10}(-x)}=\log_{10}\sqrt{x^2} }$

$\small\text{ \longrightarrow\sqrt{\dfrac{\log(-x)}{\log10}}=\dfrac{\log|x|}{\log10} }$

First let us find domain of the equation.

The restrictions are,

• $\small-x\in(0,\ \infty)\quad\dots(i)$

• $\small\dfrac{\log(-x)}{\log10}\in\left[0,\ \infty)\quad\dots(ii)$

• $\small|x|\in\left(0,\ \infty)\quad\dots(iii)$

From (i),

$\small\longrightarrow x\in(-\infty,\ 0)\quad\dots(1)$

From (ii),

$\small\longrightarrow\log(-x)\in\left[0,\ \infty)$

$\small\longrightarrow-x\in\left[1,\ \infty)$

$\small\longrightarrow x\in\left(-\infty,\ -1]\quad\dots(2)$

From (iii),

$\small\longrightarrow x\in\mathbb{R}-\{0\}\quad\dots(3)$

Taking (1) ∧ (2) ∧ (3),

$\small\longrightarrow x\in\left(-\infty,\ -1]$

This is the domain. From this we can see $\smallx<0$ so $\small|x|=-x.$

Then our equation becomes,

$\small\text{ \longrightarrow\dfrac{\sqrt{\log(-x)}}{\sqrt{\log10}}=\dfrac{\log(-x)}{\log10} }$

$\small\longrightarrow\sqrt{\log10\cdot\log(-x)}=\log(-x)$

$\small\longrightarrow\log10\cdot\log(-x)=\log^2(-x)$

$\small\longrightarrow\log^2(-x)-\log10\cdot\log(-x)=0$

$\small\longrightarrow\log(-x)\big(\log(-x)-\log10\big)=0$

$\small\Longrightarrow\log(-x)\in\{0,\ \log10\}$

$\small\longrightarrow-x\in\{1,\ 10\}$

$\small\text{ \longrightarrow\underline{\underline{x\in\{-10,\ -1\}}} }$

So there are exactly two solutions. Hence (c) is the answer.

Answer 2

Answer:

Question :-

• $\sqrt{ log( - x) } = log_{10}( \sqrt{x {}^{2} } )$
• is,

1. none
2. exactly 1
3. exactly 2.
4. 4.

♧Answer :-

• Here we get given question answer as number of real solution of a equation has exactly 2 solutions.

♧Given :-

• Here given some real solution that is ,

$\sqrt{ log( - x) } = log_{10}( \sqrt{ {x}^{2} } )$

□To find :-

• The number of real solution.

♧Solution :-

• Here refer the given attachment for better understanding.
• It helps you a lot.

Hope it helps u mate.

$\mathcal\purple{Thank you .}$