Question

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

Answer 1

Answer:

x = -1

x = -10

exactly 2 solutions

Step-by-step explanation:

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

Let say

-x = y  Where y is +ve  ( as log of -ve is undefined)

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

√(-y)² = √y² = ±y but only +y is taken as log of -ve is undefined

$\sqrt{log_{10}(y)} = log_{10}{(y)} }$

Let say

$\sqrt{log_{10}(y)} = z$

z = z²

=>z² - z = 0

=> z(z-1) = 0

=> z = 0  or z = 1

$\sqrt{log_{10}(y)} = 0 \ or \ 1$

Squaring both sides

$log_{10}{(y)} = 0 \ or \ 1$

y = 1  or 10

x = -y    

x = -1

x = -10

exactly 2 solutions

Answer 2

Answer: c) Exactly 2

Step-by-step explanation:

Given:

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

$=>\sqrt{log_{10}(-1)+log_{10} (x)}=2log_{10}x$

$Let:log_{10}(x) = u$

$=>\sqrt{log_{10}(-1) + u} = 2u$

On Squaring both sides, we get

$=>log_{10}(-1) + u = 2u^2$

Undefined

From the above solution We came to conclusion that there are no real soltuions for this equation.

This is wrong method of solving this question, which often happens (Explained Later)

Now solving it by another method:-

Given,

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

We know that,  log is not defined for negative values. So here -x must be  a positive value for real solutions.

Now, Putting -x = y

$\sqrt{\log_{10}{y}}=\log_{10}{\sqrt{(-y)^2}}\\\;\\\sqrt{\log_{10}{y}}=\log_{10}{\sqrt{y^2}}\\\;\\\sqrt{\log_{10}y}=\log_{10}y\\\;\\\text{Puitting}\;\log_{10}{y}=k\;\text{in above equation}\\\;\\\sqrt{k}=k\\\;\\\text{Making square of both sides}\\\;\\k=k^2\\\;\\k^2-k=0\\\;\\k(k-1)=0\\\;\\k=0\;\;or\;\;k=1$

When,  k = 0

$\log_{10}y=0\\\;\\\log_{10}y=\log_{10}1\\\;\\y=1\\\;\\-x=1\\\;\\x=-1$

When, k = 1

$\log_{10}y=1\\\;\\\log_{10}y=\log_{10}10\\\;\\y=10\\\;\\-x=10\\\;\\x=-10$

Note:-

There is a difference between  $2\log(-x)\;\;and\;\;\log{x^2}$

Let us take an example,

If We say ,

$\log{x^2}=\log25\\\;\\\text{Then solutions willl be}\\\;\\x=5\;\;or\;\;x=-5$

If We say,

$2\log(-x)=\log25\\\;\\\text{Then the solution will be only}\\\;\\x=5$

Here we didn't consider -5 as a solution because if we put x = -5 in second case we get a undefined result,

As conclusion we can say, Basically They are same, But Practically they are different.

Also If log is  defined as a negative variable , its ok because It may be two possibilities either variable is  negative or positive. Although positive value tends to be undefined. But What about negative. Yes! Its accepetable.