I know that \(2^-2 = 1/4\), but how do you go from \(2^x/4 to 2^(x-2)\)

Make sure to follow posting guidelines. Topic title must be reflective of the first few words of the question itself.

As for the question,

\(2^x - 2^{x-2}=3*2^{13}\) ---> \(2^x - 2^x/4=3*2^{13}\)

--->\((3/4)*2^x = 3*2^{13}\) --->\(2^x = 2^{13}*4 = 2^{13}*2^2 = 2^{15}\)

Thus, comparing both sides of the equation, you get, x=15.

D is thus the correct answer.

You can also use the options to directly substitute the values and see which one fits the given equation.

Hope this helps.

Another way to solve this is for example try doing 2^X-2^X-1 = 3(2^3) you will get answer as x=5, so similarly in the above question it should be 13+2 = 15
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\(2^x-2^{(x-2)}=3*2^{13}\)

\(2^{(x-2)} (2^2 - 1) = 3*2^{13}\)

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

Therefore,

\(x - 2 = 13\)

\(x = 15\)

(D)
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Hi pushpitkc, can you please explain some steps below ?

\(2^x-2^{x}*2^{-2}=3*2^{13}\) how do we additional 2 on the left side ?

\(2^x-\frac{2^{x}}{2^{2}}=3*2^{13}\) how do we get fraction on the left side ?

\(2^x(1-\frac{1}{2^{2}})=3*2^{13}\) how after this

\(2^x(\frac{2^2-1}{2^{2}})=3*2^{13}\) we get this on the left side

\(2^x*\frac{3}{2^{2}}=3*2^{13}\) and how we get this on the left side ?

\(2^x=2^{15}\) hopefully this step will be clear after the previous one

\(x=15\).

many thanks
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pushpitkc,
thanks for explanation, i somehow dont get based on which formula did you solve the below expression, (you provide all formulas but i couldnt recognize the pattern (i mean which formula to apply)
Here, \(2^{x-2} = 2^{x + (-2)} = 2^x*2^{-2}\)
can you please explain somehow
thanks and have a good weekend

generis perhaps you can explain from different angle if such angle exists ? I chase quant like dog chases its tale … trying to "catch" understanding ..., having hard time dealing with exponents now hope are enjoying the weekend too
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In English I speak with a dictionary, and with people I am shy.

Tenacity, thy name is dave13

I will solve a different way, in two parts.

In Part II, I will try to explain how or why a rule is working.

You are fairly warned here:
Sometimes, if we get too immersed in a specific problem,
we get stuck. Maybe if we "start over" it will help.

I. Solving without much explanation (see Part II)

\(2^x-2^{(x-2)}=3*2^{13}\): "Break apart" LHS's second term

\(2^{x}-(2^{x}*2^{-2})=3*2^{13}\): Rewrite the term that has a negative exponent

\(2^{x}-(2^{x}*\frac{1}{2^2})=3*2^{13}\): LHS parentheses: combine into one term

\(2^{x}-(\frac{2^{x}}{2^2})=3*2^{13}\): Clear the fraction

\(2^x*(2^2)-\frac{2^{x}}{2^{2}}*(2^2)=3*2^{13}*(2^2)=\)

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

LHS: factor out \(2^{x}\)
RHS: Add exponents of 2

\(2^{x}(2^2-1)=3*2^{(13+2)}=\)

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

Factor out 3, result is

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

Same base, different exponents: set exponents equal

\(x=15\)

FINIS

Part II - same as above, explained in detail
(a.k.a: Dear Peanut Gallery: Go away. Thanks.)

Original: \(2^x-2^{(x-2)}=3*2^{13}\)

Let's start with LHS: \(2^x-2^{(x-2)}\)

1) Rewrite LHS in two ways

A) Rewrite the second term:
\(2^{(x-2)}=2^{x}*2^{-2}\)

The product rule \(a^{n}a^{m}a^{q}=a^{n+m+q}\) . . .
from the other direction

\(a^{n+m+q}=a^{n}a^{m}a^{q}\) AND
\(a^{(n-m)} = a^{n + (-m)}=a^{n}a^{-m}\)

Adding a negative number is the same as subtracting.
IMO, this one is a black hole. Let it go. Use the practical tip.

Practical tip: If we have a number raised to a bunch of different
powers all in the same exponent, just "break up" that long exponent.

Write a 2, and the exponent, for each exponent. Watch the sign.
If it helps, put a multiplication sign in between each 2

Example
\(2^{(a+b+2-c)}=\)

\(2^{a}*2^{b}*2^2*2^{-c}\)

So we just "unpacked" \(2^{x-2}\) the same way

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

B) Now rewrite just this little term: \(2^{-2}\)

\(2^{-2}=\frac{1}{2^2}\)

That rewrite comes from the negative exponent rule, discussed here

C) Put together the two LHS rewrites: \(2^{x}*\frac{1}{2^2}\)

Multiply to make one term: \(=\frac{2^{x}}{2^2}\)

Write "new" LHS in full: \(2^x-\frac{2^{x}}{2^2}\)

4) Substitute new LHS into the original

\(2^x-\frac{2^{x}}{2^2}=3*2^{13}\)

5) Clear the fraction. Multiply each term by \(2^2\)

\(2^x*(2^2)-\frac{2^{x}}{2^{2}}*(2^2)=3*2^{13}*(2^2)\)

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

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

(Just no explicit multiplication signs IN terms ... Often they are not written)

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

Factor out \(2^{x}\)
\(2^{x}2^2-2^{x}=\)
\(2^{x}*(2^2-1)=\)
\(2^{x}*(4-1)=\)
\(2^{x}*(3)\)

That factoring is the same as this kind of factoring:

\((ax-a)=\)
\((a*x)-(a*1)=\)
\(a*(x-1)=\)
\(a(x-1)\)

Here is good information on factoring.
RESULT, LHS \(2^{x}*(3)\)

8) RHS: add exponents of the two 2s
RHS = \(3*2^{13}2^{2}=\)

\(3*2^{(13+2)}=3*2^{15}\)

An example of the exponent product rule.
Same base, different exponents, multiplication of bases?
Add the exponents.

Here is an explanation of the "product" rule and more.
More information to help.

RHS result:\(3*2^{15}\)

7) Combine LHS from #6 and RHS from #7:

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

8) Factor out ("cancel") the 3s:

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

9) Set exponents equal = answer

\(x=15\)

Hope that helps.

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In the depths of winter, I finally learned
that within me there lay an invincible summer.
-- Albert Camus, "Return to Tipasa"

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

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.

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 subtraction
If 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.

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

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.

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.

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.
\(2^{x}*(3)=(3)*2^{15}\)
\(\frac{2^{x}*(3)}{(3)}=\frac{(3)*2^{15}}{(3)}\)
\(2^{x}*1=2^{15}*1\)
\(2^{x}=2^{15}\)
\(x=15\)

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"