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

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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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New post 04 Mar 2015, 06:55
Zarrolou wrote:
I can explain that:

\(\frac{2^x}{4}=2^1^3=2^x^-^2=2^1^3\)

Hope now it's clear


I know that \(2^-2 = 1/4\), but how do you go from \(2^x/4 to 2^(x-2)\)
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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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New post 04 Mar 2015, 06:57
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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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New post 26 Apr 2015, 19:56
Another way to do the math I believe...

Step 1: bring the exponent (x-2) down: 2^x - 2^x /2^2 = 3*(2^13)
Step 2: factor out the 2^x: 2^x(1-1/2^2)=3*(2^13)
Step 3: solve inside the parentheses: 2^x(3/4)=3*(2^13)
Step 4: divide by 3/4: 2^x=3*(2^13)*4/3
Step 5: cancel out the 3's 2^x=2^13*4
Step 6: change 4 into exponent: 2^x=2^13*2^2=2^15

Another great method is also just testing the answer choices - I was lucky and started with 15 first so didnt take too long!
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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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New post 09 Jan 2016, 10:35
pepo wrote:
\(2^x - 2^{x-2}=3*2^{13}\), what is the value of x?

A 9
B 11
C 13
D 15
E 17


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.
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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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New post 09 Jan 2016, 10:36
1
pepo wrote:
\(2^x - 2^{x-2}=3*2^{13}\), what is the value of x?

A 9
B 11
C 13
D 15
E 17



\(2^x - 2^{x-2}=3*2^{13}\) can be written as..
\(2^x - 2^x/2^2=3*2^{13}\)..
\(2^x(1-1/4)=3*2^{13}\).
\(3* 2^{x-2}=3*2^13\)....
or x-2=13..
x=15
ans D
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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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New post 25 Aug 2017, 22:14
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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Re: If 2^x-2^(x-2)=3*2^13 what is the value of x?  [#permalink]

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New post 28 Feb 2018, 22:28
Hi All,

Factoring the equation is a useful approach here. You can actually take it a step further by TESTing THE ANSWERS (you might find it easier to manipulate numbers than to manipulate variables). One of those numbers IS the value of X, so you can plug them in, do the necessary math and find the one value that balances out the equation. Here's how to approach it:

Since the "right side" of the equation is greater than 2^14, X must be bigger than 14 (since the "left side" of the equation involves subtraction). So we can eliminate A, B and C.

Let's TEST answer D: 15

If X = 15, then...

2^15 - 2^13 can be factored into...
(2^13)(2^2 - 1) =
(2^13)(3)

This is EXACTLY what's on the "right side" of the equation, so X MUST be 15

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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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New post 21 Mar 2018, 03:14
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


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

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

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

Therefore,

\(x - 2 = 13\)

\(x = 15\)

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

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New post 22 Mar 2018, 15:45
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



We are given the equation 2^x – 2^(x-2) = 3(2^13) and we must determine x. We can start by rewriting the equation as follows:

2^x – (2^x)(2^-2) = 3(2^13)

Next we can factor out 2^x from the left hand side of the equation.

2^x(1 – 2^-2) = 3(2^13)

2^x(1 – 1/4) = 3(2^13)

2^x(3/4) = 3(2^13)

Multiplying both sides by 4/3 gives us:

2^x = 3(2^13) * 4/3

2^x = (2^13) * 4

2^x = 2^13 * 2^2

2^x = 2^15

x = 15

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

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New post 14 Jun 2018, 08:07
Bunuel wrote:
TheGerman wrote:
Zarrolou wrote:
I can explain that:

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



Hey ! Unfortunately, I still don't get where that '3' comes from... Just don't see it... Could anyone explain further in more detail?! Would be great!


Step by step:
\(2^x-2^{x-2}=3*2^{13}\) after this

\(2^x-2^{x}*2^{-2}=3*2^{13}\) 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 ? :? and how after this

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

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

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

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

\(x=15\).

Hope it's clear.




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

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New post 14 Jun 2018, 11:11
1
dave13 wrote:
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 :)



Hi dave13

Some formulae in exponents that you must remember in exponents

\(a^0 = 1\) | \(a^1 = a\) | \(\sqrt{a} = a^{\frac{1}{2}}\) | \(a^{-n} = \frac{1}{a^n}\)

\(a^{m+n} = a^m*a^n\) | \(\frac{a^m}{a^n}=a^{m−n}\) | \(a^n = \frac{1}{a^{-n}}\)

Coming back to the problem at hand

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

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

\(2^x(1-\frac{1}{2^{2}})=3*2^{13}\) Here, we take \(2^x\) as common from both the expressions

\(2^x(\frac{2^2-1}{2^{2}})=3*2^{13}\) Here, we simplify \(1 - \frac{1}{2^2}\) as \(\frac{2^2 - 1}{2^2} = \frac{4-1}{2^2} = \frac{3}{2^2}\)

\(2^x*\frac{3}{2^{2}}=3*2^{13}\) -> \(2^x=2^{15}\) Here, we take \(2^2\) from left hand side to right hand side
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If 2^x-2^(x-2)=3*2^13 what is the value of x?  [#permalink]

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New post 16 Jun 2018, 03:59
pushpitkc wrote:
dave13 wrote:
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 :)



Hi dave13

Some formulae in exponents that you must remember in exponents

\(a^0 = 1\) | \(a^1 = a\) | \(\sqrt{a} = a^{\frac{1}{2}}\) | \(a^{-n} = \frac{1}{a^n}\)

\(a^{m+n} = a^m*a^n\) | \(\frac{a^m}{a^n}=a^{m−n}\) | \(a^n = \frac{1}{a^{-n}}\)

Coming back to the problem at hand

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

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

\(2^x(1-\frac{1}{2^{2}})=3*2^{13}\) Here, we take \(2^x\) as common from both the expressions

\(2^x(\frac{2^2-1}{2^{2}})=3*2^{13}\) Here, we simplify \(1 - \frac{1}{2^2}\) as \(\frac{2^2 - 1}{2^2} = \frac{4-1}{2^2} = \frac{3}{2^2}\)

\(2^x*\frac{3}{2^{2}}=3*2^{13}\) -> \(2^x=2^{15}\) Here, we take \(2^2\) from left hand side to right hand side



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

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New post 16 Jun 2018, 13:18
1
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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New post 16 Jun 2018, 19:59
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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 \(2^{x}\)
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. :-D
:dazed :dazed :fingers_crossed:
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If 2^x-2^(x-2)=3*2^13 what is the value of x?  [#permalink]

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New post 18 Jun 2018, 04:15
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 \(2^{x}\)
RHS: Add exponents of 2
\(2^{x}(2^2-1)=3*2^{(13+2)}=\)
\(2^{x}*(3)=(3)*2^{15}\)

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

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New post 08 Jul 2018, 19:00
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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 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 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? :grin: 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. :thumbdown: :thumbdown: :thumbdown: :thumbdown: :thumbdown:
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. :fingers_crossed:
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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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New post 12 Aug 2018, 23:14
Can also test answer choices coupled w/ intuition to quickly get to an answer if the algebra / rearrange route doesn't automatically come to mind (or if you're prone to making stupid/sloppy mistakes like myself!)...

We know that 2^x - 2^x-2 = 3(2^13) ---> hence we need to be able to factor out at least 2^13 from the LHS but that can only be the case if both terms have at least 2^13...b/c the 2nd term is x-2, we know that we need x to be at least 15

Quick check to confirm --> 2^15 - 2^13 --> 2^13(2^2 - 1) = 2^13(4 - 1) = 2^13(3)

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

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New post 10 Mar 2019, 10:04
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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New post 10 Mar 2019, 12:21
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.

I wrote extensively above about this issue.

(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 \(2^{x}\)
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

\(x=15\)


I hope that analysis helps.
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If 2^x-2^(x-2)=3*2^13 what is the value of x?   [#permalink] 10 Mar 2019, 12:21

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