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705-805 (Hard)|   Exponents|      
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Bunuel
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If 5^x – 3^y = 13438 and 5^x-1 + 3^y+1 = 9686, then what is the value 13 Apr 2020, 07:32
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D
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95% (hard)

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52% (03:04) correct 48% (03:12) wrong based on 161 sessions

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If \(5^x – 3^y = 13438\) and \(5^{x-1} + 3^{y+1} = 9686\), then what is the value of \(x+y\) ?

A. 10
B. 11
C. 12
D. 13
E. 14


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D
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If 5^x – 3^y = 13438 and 5^x-1 + 3^y+1 = 9686, then what is the value 13 Apr 2020, 07:58
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Bunuel
If \(5^x – 3^y = 13438\) and \(5^{x-1} + 3^{y+1} = 9686\), then what is the value of \(x+y\) ?

A. 10
B. 11
C. 12
D. 13
E. 14


Are You Up For the Challenge: 700 Level Questions
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The first expression given to us is:
    • \(5^x – 3^y = 13438 ……(i)\)
The second expression given to us is:
    • \(5^{x-1} + 3^{y+1} = 9686\)
    • \(5^x + 15* 3^y = 9686*5\)
    • \(5^x + 15 * 3^y= 48430…(ii)\)
Subtracting expression (i) from equation (ii), get
\(5^x + 15 * 3^y= 48430…(ii)\)
\(5^x – 3^y = 13438 ……(i)\)
Subtracting…………………………….
\(16* 3^y = 48430 – 13438\)
\(16 * 3^y = 34992\)
\(3^y = 2187\)
\(3^y = 3^7\)
Thus, \(y = 7\)
On substituting the value of y in equation (i), we get
\(5^x – 2187 = 13438\)
\(5^x = 15625\)
\(5^x = 5^6\)
\(x = 6\)
Thus, \(x + y = 6 + 7 = 13\)

Correct answer is Option D
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If 5^x – 3^y = 13438 and 5^x-1 + 3^y+1 = 9686, then what is the value 13 Apr 2020, 08:05
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\(5*5^{x-1} - 3^y = 13438\)

\(5^{x-1} + 3*3^y = 9686\)

\(5^{x-1} = a\) and \(3^y=b\)

5a-b=13438....(1)
a+3b=9686....(2)

From (1) and (2)

a=3125 or 5^5; b= 2187 or 3^7

x = 6 and y =7
x+y=13


approach 2- Bit quicker than first one.

Unit digit of 3^y is 7; hence y is in the form of 4k+3. y can be 3 or 7. [3^11>9686]

When you put y=3, x is not an integer(We can easily reject this one.)

Put y=7; we get x=6

Bunuel
If \(5^x – 3^y = 13438\) and \(5^{x-1} + 3^{y+1} = 9686\), then what is the value of \(x+y\) ?

A. 10
B. 11
C. 12
D. 13
E. 14


Are You Up For the Challenge: 700 Level Questions
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If 5^x – 3^y = 13438 and 5^x-1 + 3^y+1 = 9686, then what is the value 13 Apr 2020, 08:13
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Bunuel
If \(5^x – 3^y = 13438\) and \(5^{x-1} + 3^{y+1} = 9686\), then what is the value of \(x+y\) ?

A. 10
B. 11
C. 12
D. 13
E. 14


Are You Up For the Challenge: 700 Level Questions
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I believe I made it quite simpler while solving

I took just one equation \(5^x – 3^y = 13438\)

5^x will always have unit digit 5
and 5^x – 3^y has unit digit 8
i.e. something subtracted from 5 should give unit digit 8

then that something must be 7 (because 15-7 = 8)

i.e. 3^y must have unit digit 7

Now lets do some trial and error

\(5^3 = 125\)
\(5^4 = 625\)
\(5^5 = 3125\)
\(5^6\) = approx 15000 with unit digit 5 (we wanted to reach close to 13438 which we have)

\(3^4 = 81\)
\(3^5 = 243\)
\(3^6 = 729\)
\(3^7\) = approx 2100 with unit digit 7 (we wanted to reach close to 15000-13438≈2000 which we have)

i.e. \(5^x – 3^y = 13438\) resembles with \(5^6 – 3^7 = 13438\)

i.e. x = 6 and y = 7

I verified these values with second equation and they satisfied on unit digit calculation so BINGO!

x+y = 6+7 = 13

Answer: Option D

Video Solution attached here for better understanding.

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If 5^x – 3^y = 13438 and 5^x-1 + 3^y+1 = 9686, then what is the value 13 Apr 2020, 08:49
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5^x–3^y=13438
5^(x−1)+3^(y+1)=9686

let, 5^x =a, 3^y = b
solving, a=15625, b= 2187
or, x= 6, y=7

so, x+y = 13 .

correct answer D
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If 5^x – 3^y = 13438 and 5^x-1 + 3^y+1 = 9686, then what is the value 22 May 2020, 08:08
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Bunuel
If \(5^x – 3^y = 13438\) and \(5^{x-1} + 3^{y+1} = 9686\), then what is the value of \(x+y\) ?

A. 10
B. 11
C. 12
D. 13
E. 14
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\(5^x= 13438 + 3^y = a+3^y\)

\(a+3^y/5 + 3^y*3 = 9686,\)
\(3^y+ 3^y*3*5 = 9686*5-a,\)
\(3^y(16) = 9686*5-a,\)
\(3^y = (9686*5-13438)/16,\)
\(3^y = 2187,…y=7\)

\(5^x= 13438 + 2187,\)
\(5^x = 15625,…x=6\)

\(x+y=13\)

ans (D)
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If 5^x – 3^y = 13438 and 5^x-1 + 3^y+1 = 9686, then what is the value 30 Aug 2020, 01:03
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Bunuel
If \(5^x – 3^y = 13438\) and \(5^{x-1} + 3^{y+1} = 9686\), then what is the value of \(x+y\) ?

A. 10
B. 11
C. 12
D. 13
E. 14


Are You Up For the Challenge: 700 Level Questions

The first expression given to us is:
    • \(5^x – 3^y = 13438 ……(i)\)
The second expression given to us is:
    • \(5^{x-1} + 3^{y+1} = 9686\)
    • \(5^x + 15* 3^y = 9686*5\)
    • \(5^x + 15 * 3^y= 48430…(ii)\)
Subtracting expression (i) from equation (ii), get
\(5^x + 15 * 3^y= 48430…(ii)\)
\(5^x – 3^y = 13438 ……(i)\)
Subtracting…………………………….
\(16* 3^y = 48430 – 13438\)
\(16 * 3^y = 34992\)
\(3^y = 2187\)
\(3^y = 3^7\)
Thus, \(y = 7\)
On substituting the value of y in equation (i), we get
\(5^x – 2187 = 13438\)
\(5^x = 15625\)
\(5^x = 5^6\)
\(x = 6\)
Thus, \(x + y = 6 + 7 = 13\)

Correct answer is Option D
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Thanks for this, can you please explain how do you get 16?
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If 5^x – 3^y = 13438 and 5^x-1 + 3^y+1 = 9686, then what is the value Updated on: 30 Aug 2020, 12:02
Originally posted by GMATGuruNY on 30 Aug 2020, 04:25.
Last edited by GMATGuruNY on 30 Aug 2020, 12:02, edited 2 times in total.
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Bunuel
If \(5^x – 3^y = 13438\) and \(5^{x-1} + 3^{y+1} = 9686\), then what is the value of \(x+y\) ?

A. 10
B. 11
C. 12
D. 13
E. 14
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Subtract the second equation from the first:
\((5^x – 3^y) - (5^{x-1} + 3^{y+1}) = 13438-9686\)

Combine like terms:
\((5^x - 5^{x-1}) - (3^y + 3^{y+1}) = 3652\)

Factor out the smaller exponent from the first two terms (x-1) and the smaller exponent from the last two temrs (y):
\(5^{x-1}(5-1) - 3^y(1+3) = 3652\)

Simplify:
\(5^{x-1}(4) - 3^y(4) = 3652\)
\(5^{x-1} - 3^y = 938\)

List consecutive powers of 5 and consecutive powers of 3 and search for a combination with a difference of 938:
5, 25, 125, 625, 3125...
3, 9, 27, 81, 243, 729, 2187...
The values in green yield a difference of 938:
\(3125-2187 = 938\)

Thus:
\(5^{x-1} = 3125\)
\(5^{x-1} = 5^5\)
\(x-1 =5\)
\(x=6\)

\(3^y = 2187\)
\(3^y = 3^7\)
\(y=7\)

\(x+y = 6+7 = 13\)

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D
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If 5^x – 3^y = 13438 and 5^x-1 + 3^y+1 = 9686, then what is the value 30 Aug 2020, 08:30
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If \(5^x – 3^y = 13438\) and \(5^{x-1} + 3^{y+1} = 9686\), then what is the value of \(x+y\) ?

A. 10
B. 11
C. 12
D. 13 --> correct
E. 14

Solution:
\(5^x – 3^y = 13438\) --> multiply both side by 3
\(3*5^x – 3^{y+1} = 3*13438=40314 ....(i)\)
\(5^{x-1} + 3^{y+1} = 9686 ....(ii)\)
Adding (i) & (ii) =>
\(5^x + 5^{x-1} = 40314 + 9686 = 50000\)
=> \(3*5^x + 5^{x-1} = 40314 + 9686 \)
=> \(5^x(3 + \frac{1}{5}) = 50000\)
=> \(5^x * \frac{16}{5} = 50000 = 5*10000\)
=> \(5^{x-2} = 50000 = 25*25=5^{4}\)
=> \(x=6\)

Now substituting x=6 in equation (ii)
\(5^5 + 3^{y+1} = 9686\)
=> \(3^{y+1} = 9686-5^5=9686-3125=6561=3^8\)
=> \(y=7\)
So => \(x+y=6+7=13\)
Answer: D
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If 5^x – 3^y = 13438 and 5^x-1 + 3^y+1 = 9686, then what is the value 30 Aug 2020, 09:03
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Asked: If \(5^x – 3^y = 13438\) and \(5^{x-1} + 3^{y+1} = 9686\), then what is the value of \(x+y\) ?

\(5*5^{x-1} + 5*3^{y+1} = 5*9686 = 48430\)
\(5^x + 15*3^y = 48430\)
\(5^x – 3^y = 13438\)
\(14*3^y = 34992\)
\(3^y = 2500\)
y = 7.12

\(5^x = 13438 - 2500 = 10938\)
x = 5.78

x + y = 7.12 + 5.78 = 13

IMO D
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If 5^x – 3^y = 13438 and 5^x-1 + 3^y+1 = 9686, then what is the value 26 Sep 2025, 02:33
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