# Aptitude - Surds and Indices - Discussion

Discussion Forum : Surds and Indices - General Questions (Q.No. 2)

2.

If | a |
x - 1 |
= | b |
x - 3 |
, then the value of x is: |
||||

b |
a |

Answer: Option

Explanation:

Given | a |
x - 1 |
= | b |
x - 3 |
|||

b |
a |

a |
x - 1 |
= | a |
-(x - 3) |
= | a |
(3 - x) |
|||||||

b | b | b |

*x* - 1 = 3 - *x*

2*x* = 4

*x* = 2.

Discussion:

26 comments Page 1 of 3.
Nisikanta said:
1 decade ago

(a/b)^x-1= (b/a)^x-3

* Remember, sinplification is always solved by 2 or more fraction or number or alphabet.

:If the alphabet fraction is not equal to one anthor,we have reranged it.

= (a/b)^x-1= (a/b)^-(x-3)= (a/b)^(3-x)

: Then we have to take power for simplification.

= x-1= 3-x

* When the number goes after or before '=' sign the sign changes

: Now we have to change the places.

= 2x=4

=x=2

* Remember, sinplification is always solved by 2 or more fraction or number or alphabet.

:If the alphabet fraction is not equal to one anthor,we have reranged it.

= (a/b)^x-1= (a/b)^-(x-3)= (a/b)^(3-x)

: Then we have to take power for simplification.

= x-1= 3-x

* When the number goes after or before '=' sign the sign changes

: Now we have to change the places.

= 2x=4

=x=2

Husain SR said:
8 years ago

Here (a/b)^x-1 and (b/a)^x-3 now if we reciprocal b/a to a/b then there is negative sign on it.

(a/b)^-(x-3).

Let compare thier powers;

x - 1 = -(x - 3).

x - 1 = -x + 3

x + x = 1 + 3.

2x = 4

x = 2 is answer.

(a/b)^-(x-3).

Let compare thier powers;

x - 1 = -(x - 3).

x - 1 = -x + 3

x + x = 1 + 3.

2x = 4

x = 2 is answer.

(1)

Manu said:
9 years ago

The problem can be solved by using the property "If the bases are equal then there powers must be equal". So in order to find the value we have to make the bases equal, after equating solve for x.

URVASHI said:
4 years ago

Simply Cross Multiply them a with a and b with b.

It becomes;

[a ^(x-1) ][a^(x-3)] = [b^(x-1)][b^(x-3)].

Power will get add on both sides.

Then,

a^(2x-4) = b^(2x-4),

Simply, x=2.

It becomes;

[a ^(x-1) ][a^(x-3)] = [b^(x-1)][b^(x-3)].

Power will get add on both sides.

Then,

a^(2x-4) = b^(2x-4),

Simply, x=2.

(2)

Rakesh KIIT said:
1 decade ago

(a/b)^x-1=(b/a)^x-3

can be written as :

After cross multiplication :

(a)^(2x-4)=(b)^(2x-4)

=>(a/b)^(2x-4)=1

=>(a/b)^(2x-4)=(a/b)^0

=> 2x-4=0

=>x=4/2

=>x=2

can be written as :

After cross multiplication :

(a)^(2x-4)=(b)^(2x-4)

=>(a/b)^(2x-4)=1

=>(a/b)^(2x-4)=(a/b)^0

=> 2x-4=0

=>x=4/2

=>x=2

Shweta said:
7 years ago

@Husain.

When you compare the power then , x-1=-x-3.

Then how you change the sign of ,x-1=x+3?

When you compare the power then , x-1=-x-3.

Then how you change the sign of ,x-1=x+3?

SANJIV KUMAR said:
1 decade ago

Everybody can you tell me that how cross multiplication (a)^(2x-4) = (b)^(2x-4) in this answer.

Ravi kiran said:
1 decade ago

@pooja

(b/a) can be written as (a/b)^-1.

Based on this concept this problem is solved.

(b/a) can be written as (a/b)^-1.

Based on this concept this problem is solved.

Heman said:
7 years ago

(a/b)^x-1=(b/a)x-3.

=>a^x/b^1=a^-3/b^-x,

=>a/b^x-1=a/b^3-x,

=>x-1=3-1,

=>x=2.

=>a^x/b^1=a^-3/b^-x,

=>a/b^x-1=a/b^3-x,

=>x-1=3-1,

=>x=2.

Nem prasad said:
1 decade ago

Can anybody tell me what is the concept through which this problem has been solved ?

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