dave13 wrote:

generis many thanks for your explanation

just a few questions

dave13 , I will break your questions up.

First I want to mention an excellent post by

mikemcgarry Adding and Subtracting Powers on the GMAT. It should make this problem (and, I hope, my answer) easier to understand.

I think variables, exponents and strange fractions might be making things seem really chaotic. These issues keep cropping up (for a lot of people):

• Exponents do not distribute over addition and subtraction.

• We cannot use the "same base" rule if there is addition or subtraction between terms.

• Factoring. Most people easily understand distributing. The other direction is harder.

Distributing: \(a(b + c) = ab + ac\)

Factoring: \(ab + ac = a(b + c)\)

**Quote:**

Question # 1: here \(2^x*(2^2)-\frac{2^{x}}{2^{2}}=3*2^{13}*(2^2)\)

when you clear fraction on the RHS why \((2^2)\)you only multiply by \(2^{13}\) and not by both numbers including 3 ? like this RHS -- > \(3*(2^2) *2^{13}*(2^2)\)

To clear this fraction, we multiply each

single term by 2. LHS has TWO terms separated by a subtraction operator.

RHS has only ONE term, which consists of two factors multiplied. The numbers are not separated by an addition or subtraction sign.

CORRECT - Multiplying RHS by \(2^2\) only once because RHS is a SINGLE term

\((3)*(2^{13})\)

\((3)*(2^{13})*(2^2)=\)

\((3)*(8,192)*(4)=98,304\)

INCORRECT - Multiplying both factors of RHS by \(2^2\), as if RHS were TWO terms

\(3*2^{13}\)

\((3*2^2)*(2^{13}*2^2)=\)

\((3*4)*(8,192)*(4)=\)

\((12)*(32,768)=393,216\)

\(393,216\) is FOUR times the correct value.

I incorrectly multiplied a single, individual term by \(2^2\) twice.

**Quote:**

Question # 2: here \(2^{x}2^2-2^{x}=3*2^{13}2^{2}\) when when we factor out \(2^x\)

LHS: factor out \(2^{x}\)

i thought from here \(2^{x}2^2-2^{x}\) we simply add exponents like this

\(2^{2+x} - 2^x\) so here i equate bases and i get \(2+x-x\)

but then i get only 2, since x`s are cancelled out

shouldnt we equate bases ?

as i tried to demonstrate above

No. You cannot equate bases until you have:

-- only multiplication or division of bases raised to exponents AND you have

-- an equals sign somewhere

Exponents do not distribute over addition and subtractionIf you see an addition or subtraction sign between bases raised to powers, the exponents cannot just "hop over" the operators.See the article by

mikemcgarry , above.

**Quote:**

On the other hand if i use your approach of factoring out \(2^x\) . . .

\(2^{x}(2^2-1)\) from here if i want to get back to initial equation (before factoring out \(2^x\))

Why would you want to get back to the initial equation?

It's the source of the

trouble!

I factor out \(2^{x}\) because

I cannot equate bases with exponents "over" the subtraction sign in the original equation! The way to eliminate the subtraction sign? Factor out something that is common to the two terms involved in subtraction.

Finding a factor in common**Quote:**

so i get \(2^{2+x} - 2^{x+1}\) is it correct ?

No, though you are

close. \(2^{x}(2^2-1)=(2^{x}*2^2) - (2^{x}*1)=2^{(2+x)}-2^{x}\)

You do not want to go back to the original equation. **Quote:**

~~now i equate bases and get 2+x-x+1 =3~~

You cannot equate bases. Above, you have two terms with a subtraction sign in between them. Won't work.

You must factor out a number or term that is common to the terms on each side of the subtraction sign.

\(2^{2+x} - 2^x=\)

\((2^2*2^{x}) - (2^{x}*1)\)

\((2^2*2^{x}) - (2^{x}*1)\)

What do the two terms have

in common? What is a factor of both terms? \(2^{x}\)

\((2^2)*\)

\((2^{x})\) \() -\)

\((2^{x})\) \(*(1)\)

\(2^{x} * (2^2-1)=\)

\(2^{x} * (4 - 1)=\)

\(2^{x} * 3\) Aha! Now we have a 3 on LHS, too. Now we are close to getting identical bases.

**Quote:**

Question # 3: here, i didnt get how from here

\(2^{x}*(3)=(3)*2^{15}\)Factor out ("cancel") the 3s: we get this \(2^{x}=2^{15}\)

you say factoring / cancelling out but 3 is only in bracketsand not like canceled as for example \(\frac{3}{4}\)*\(\frac{4}{7}\) here i cancel out 4s

What you say is a

common misconception.

Cancelling IS factoring out. Cancelling = DIVISION = factoring out

Your 4s do not "go away." You have reduced the fraction. You have divided the numerator and denominator by a factor of 4.

A "1" remains in the place in which the 4s were.

\((\frac{3}{4}*\frac{4}{7})=(\frac{3*4}{4*7}=(\frac{4*3}{4*7})=\)

\((\frac{4}{4}*\frac{3}{7})=(\frac{1}{1}*\frac{3}{7})=(1*\frac{3}{7})=\frac{3}{7}\)

Notice also, in your example, that there is ONLY a multiplication sign.

If we have \(\frac{3}{4}+\frac{4}{7}\) we cannot "cancel" out factors of 4.

I can indeed "cancel" just the same way. I can do so because NOW there are only multiplication signs between the numbers.

\(2^{x}*(3)=(3)*2^{15}\)

\(\frac{2^{x}*(3)}{(3)}=\frac{(3)*2^{15}}{(3)}\)

\(2^{x}*1=2^{15}*1\)

\(2^{x}=2^{15}\)

\(x=15\)

**Quote:**

well these are my questions

I would appreaciate if you could explain

If this answer does not help, you might be too close to the material.

That is, sometimes we think about things

too much.

I sincerely hope this helped.

_________________

In the depths of winter, I finally learned

that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"