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If 2^x-2^(x-2)=3*2^13 what is the value of x?  [#permalink]

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dave13 wrote:

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 Hey dave13

$$2^{x-2} = 2^{x + (-2)}$$ We do this step in order to convert the exponent into $$a^{m+n}$$

$$2^{x + (-2)} = 2^x*2^{-2}$$ Now, we use the formula - $$a^{m+n} = a^m * a^n$$

$$2^x*2^{-2} = 2^x * \frac{1}{2^{2}}$$ Now, we use the formula - $$a^{-m} = \frac{1}{a^m}$$

This is how $$2^x - 2^{x-2} = 3*2^{13}$$ becomes $$2^x - 2^x * \frac{1}{2^{2}} = 3*2^{13}$$

Once, we take $$2^x$$ as common, we will arrive at this: $$2^x(1-\frac{1}{2^{2}})=3*2^{13}$$

Hope the rest of the solution is clear!
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If 2^x-2^(x-2)=3*2^13 what is the value of x?  [#permalink]

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1
dave13 wrote:
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 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 $$2^{x}$$

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

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

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.

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?

Here is an explanation of the "product" rule and more.

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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If 2^x-2^(x-2)=3*2^13 what is the value of x?  [#permalink]

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generis many thanks for your explanation just a few questions 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 -- > $$3*(2^2) *2^{13}*(2^2)$$

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

LHS: factor out $$2^{x}$$
$$2^{x}(2^2-1)=3*2^{(13+2)}=$$
$$2^{x}*(3)=(3)*2^{15}$$

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 xs are cancelled out 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 $$2^x$$)

so i get $$2^{2+x} - 2^{x+1}$$ is it correct ? now i equate bases and get $$2+x-x+1 =3$$

shouldnt we equate bases ? as i tried to demonstrate above Question # 3: here, i didnt get how from here

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

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

we get this $$2^{x}=2^{15}$$

you say factoring / cancelling out but 3 is only in brackets and not like canceled as for example i would say example $$\frac{3}{4}$$*$$\frac{4}{7}$$ here i cancel out 4s well these are my questions I would appreaciate if you could explain Senior SC Moderator V
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If 2^x-2^(x-2)=3*2^13 what is the value of x?  [#permalink]

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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 xs 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 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.
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. _________________
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If 2^x-2^(x-2)=3*2^13 what is the value of x?  [#permalink]

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If the bases are the same on the left side of the equation is it fair to always pull out the common exponent and set that equal to base of the right side's exponent?

X-2 = 13 ?
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If 2^x-2^(x-2)=3*2^13 what is the value of x?  [#permalink]

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enigma123 wrote:
If $$2^x-2^{(x-2)}=3*2^{13}$$ what is the value of x?

A. 9
B. 11
C. 13
D. 15
E. 17

bhChicago wrote:
If the bases are the same on the left side of the equation is it fair to always pull out the common exponent and set that equal to base of the right side's exponent?

X-2 = 13 ?

bhChicago , slightly belated welcome to GMAT Club!

I am not sure whether you are asking:
(1) if there are ANY two bases that are identical (no matter what other numbers and operators might be
combined with one of both of the identical bases on each side) . . .
or
(2) if two or more identical bases on each side of the equation, and neither side has an operator such as
a plus (+) or minus (-), can we equate the exponents of the bases?

I think you mean #1.

Can we simply write $$x-2=13$$ from
the original equation?
$$2^x-2^{(x-2)}=3*2^{13}$$
No.

We do not have identical bases on the LHS and RHS.
-- There is no factor of 3 on the LHS.
We need a factor of 3 on the LHS, because 3*2 alters the 2 raised to an exponent.
Take a simple example. Let $$x = 5$$
$$(2^5 - 2^3) = (32 - 8) = 16$$
16 is not divisible by 3. We do not have an equality.

-- There is a minus sign on the LHS.
We cannot "pull out" $$x-2$$ from two different powers of 2 that are separated by a minus sign.
Exponents cannot "hop" over minus signs.

(1) We need a factor of 3 on the LHS

(2) It is likely that the LHS can be manipulated so that a factor of 3 exists on the LHS
(RHS has a factor of 3.
We are being set up to equate exponents.
We cannot do so until we have a base of 2 and a base of 3 on the LHS,
just as is the case on the RHS.
LHS almost certainly contains a factor of 3. We have to find the 3.
At the moment there are only 2s as bases.)

(3) Manipulate the LHS

$$2^x-2^{(x-2)}=3*2^{13}$$: "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 $$2^{x}$$

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

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

Factor out 3, result is

$$2^{x}=2^{15}$$

Same base, different exponents: set exponents equal

$$x=15$$

I hope that analysis helps.
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Re: If 2^x-2^(x-2)=3*2^13 what is the value of x?  [#permalink]

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Hey guys I just want to know how we got rid of the subtraction sign and where did the 1 pop up from?
Everything else I understand just this step seems to be consolidated in the explanations.
2^x−2^x/2^2

2^x(1−1/2^2)
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Re: If 2^x-2^(x-2)=3*2^13 what is the value of x?  [#permalink]

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nerdynotdirty wrote:
Hey guys I just want to know how we got rid of the subtraction sign and where did the 1 pop up from?
Everything else I understand just this step seems to be consolidated in the explanations.
2^x−2^x/2^2

2^x(1−1/2^2)

Factor out 2^x from $$2^x-\frac{2^{x}}{2^{2}}$$ to get $$2^x(1-\frac{1}{2^{2}})$$.

The same way after factoring m from m - m/4 you get m(1 - 1/4).
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If 2^x-2^(x-2)=3*2^13 what is the value of x?  [#permalink]

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Hey guys,

This has always been a favorite problem of mine - the first time I saw it, an instructor had blanked on how to solve it and emailed me a photo from his phone asking for help. He had excused himself from a tutoring session and needed me to explain how to solve it (correctly, of course) within a few minutes, so the pressure was on!

Bunuel's explanation is perfect (as always), but when the pressure was on and I wasn't exactly thinking about factoring, I did this instead - I looked to see if there were a pattern in the subtraction at left (2 to an exponent minus 2 to another exponent, two less) that would always produce 3*something on the right. So I did:

x = 3 and x-2 = 1
2^3 - 2^1 = 8 - 2 = 6

And 6 = 3(2^1), so I had a start.

x = 4 and x-2 = 2
2^4 - 2^2 = 16 - 4 = 12

And 12 = 3(2^2), so the pattern held

x = 5 and x-2 = 3
2^5 - 2^3 = 32 - 8 = 24

And 24 = 3(2^3), and the pattern became clear...
The operation at left was always producing 3*2^(x-2) as its answer, so if x-2 = 13, then x = 15.

Strategically, using small numbers to establish patterns works pretty well when huge numbers (like 3(2^13)) are in play, and when exponents are involved (exponents are essentially just repetitive multiplication, so there are bound to be some repetitive patterns involved). If you can factor like Bunuel did, that's a great way to go...but I'd recommend having the "prove patterns w/ small numers" ideology in your arsenal!
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