If 1/2^m + 1/2^m = 1/2^x, then x expressed in terms of m is

Question

If  \frac{1}{2^m} + \frac{1}{2^m} = \frac{1}{2^x} , then x expressed in terms of m is

A.  \frac{m}{2}

B.  m - 1

C.  m + 1

D.  2m

E.  m^2

EgmatQuantExpert · Answer

Solution

Given:  
•  \frac{1}{2^m}  +  \frac{1}{2^m}  =  \frac{1}{2^x}

To find:  
• The value of x in terms of m

Approach and Working:   
From the given equation, we can write,
 •  \frac{1}{2^m}  +  \frac{1}{2^m}  =  \frac{1}{2^x} 
Or,  \frac{2}{2^m}  =  \frac{1}{2^x} 
Or,  \frac{2^m}{2}  =  2^x 
Or,  2^{m-1}  =  2^x 
Or, m – 1 = x
Or, x = m – 1

Hence, the correct answer is option B.

Answer: B

WizakoBaskar · Answer

\frac{1}{2^m} + \frac{1}{2^m} = \frac{1}{2^x} ,

\frac{2}{2^m} = \frac{1}{2^x}

2^{(1 - m)} = 2^{(-x)}

Bases equal. Equate the indices.
1 - m = -x
or x = m - 1

Choice B.

pikolo2510 · Answer

\frac{1}{2^m} + \frac{1}{2^m} = \frac{1}{2^x}

\frac{2}{2^m} = \frac{1}{2^x}

Multiplying  2^m  throughout, we get

2^1 = 2^{(m-x)}

As bases are equal

m - x = 1

x = m - 1

Answer B

BrentGMATPrepNow · Answer

\frac{1}{2^m} = 2^{-m}

So,  \frac{1}{2^m} + \frac{1}{2^m} = 2^{-m} + 2^{-m}

= 2(2^{-m})

= (2^1)(2^{-m})

= 2^{1 - m}

We can also say that  \frac{1}{2^x} = 2^{-x}

This means we can say:  2^{1 - m} = 2^{-x}

So: 1 - m = -x

Multiply both sides by -1 to get: -1 + m = x

In other words, m - 1 = x

Answer: B

KSBGC · Answer

This one is pretty straight forward question.

(1+1)/ 2^m  =1 / 2^x 
2. 2^x  =  2^m 
 2^{x+1}  = 2^m 
x+1 = m
x = m-1

thus , B is the best option.

JeffTargetTestPrep · Answer

Combining the fractions on the left hand side, we have:

2/2^m = 1/2^x

Cross-multiplying yields:

(2)(2^x) = 2^m

Re-expressing 2 as 2^1, we can then simplify the left side:

(2^1)(2^x) = 2^m

2^(1+x) = 2^m

1 + x = m

x = m - 1

Answer: B

dave13 · Answer

hello EgmatQuantExpert how from this \frac{2}{2^m} = \frac{1}{2^x} you got this \frac{2^m}{2} = 2^x pls explain

thank you

dave13 · Answer

hey pushpitkc , how/why after this 2^{-m} + 2^{-m} we get = 2(2^{-m}) ?

shouldnt we get 4^{-m} ?? we need to add bases and not multiply

happy sunday

pushpitkc · Answer

Hi  dave13

Write this down somewhere and memorize it -  a^m*a^n = a^{m+n}

(One of the basic rules of exponents)

So,  2(2^{-m}) = 2^1*2^{-m} = 2^{1-m}

Hope this helps you!

Posted from my mobile device

dave13 · Answer

pushpitkc many thanks for explanation , but here we are adding bases 2^{-m} + 2^{-m} whereas we apply this formula when we multiply bases a^m*a^n = a^{m+n}

Bunuel · Answer

You are over-complicating this.

2^{-m} + 2^{-m}=2*2^{-m}  the same way x + x = 2x. Substitute x with 2^(-m) and you'll  2*2^{-m} . It's that simple.

PKN · Answer

Given LHS,  \frac{1}{2^m}  +  \frac{1}{2^m}  = 2^(-m) + 2^(-m) = 2*2^(-m) = 2^(1-m) ----------------(1)
Given RHS,  \frac{1}{2^x} = 2^(-x) ----------------------------------------------------- (2)

Given LHS=RHS

So  2^(1-m) = 2^(-x) 
Implies 1-m=-x

Hence x=m-1.

Ans Option B is the Correct answer.

bumpbot · Answer

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If 1/2^m + 1/2^m = 1/2^x, then x expressed in terms of m is

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