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ryguy904
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If x and y are integers and (15^x) + ___________________ = 15 Jul 2008, 11:15
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If x and y are integers and

(15^x) + [15^(x+1)]
___________________ = 15^y
4^y


what is the value of x?


2
3
4
5
Cannot Be Determined



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Maple
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If x and y are integers and (15^x) + ___________________ = 15 Jul 2008, 11:25
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let y=0

15^(x)+15^(x+1)=1
->
x is not equal to 2,3,4, or 5

E
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fatb
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If x and y are integers and (15^x) + ___________________ = 15 Jul 2008, 11:30
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It is impossible for 15^x = 4^y (15 does not have a prime factor of 2 in it)
So the answer is E
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If x and y are integers and (15^x) + ___________________ = 15 Jul 2008, 11:36
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You don't know that y=0.

Just simplify the equation:

\(\frac{15^x + 15^{x+1}}{4^y} = 15^y\)

\(15^x \left( \frac{1 + 15}{4^y} \right) = 15^y\)

\(15^x \left( \frac{4^2}{4^y} \right) = 15^y\)

\(15^x \left( \frac{1}{4^{y-2}} \right) = 15^y\)

Now, stem says that x and y are integer. This last equation can holds true only if \(y=2\) in order to get rid of the power of 4 (other than that you just have powers of 3 and 5 : you cannot have a power of 4 on the left or the right of the equation).

Then if \(y=2\) ==> \(x=2\) ==> Answer is (A)
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If x and y are integers and (15^x) + ___________________ = 15 Jul 2008, 13:03
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Oski
You don't know that y=0.

Just simplify the equation:

\(\frac{15^x + 15^{x+1}}{4^y} = 15^y\)

\(15^x \left( \frac{1 + 15}{4^y} \right) = 15^y\)

\(15^x \left( \frac{4^2}{4^y} \right) = 15^y\)

\(15^x \left( \frac{1}{4^{y-2}} \right) = 15^y\)

Now, stem says that x and y are integer. This last equation can holds true only if \(y=2\) in order to get rid of the power of 4 (other than that you just have powers of 3 and 5 : you cannot have a power of 4 on the left or the right of the equation).

Then if \(y=2\) ==> \(x=2\) ==> Answer is (A)
Show more

You are correct. For some reason I don't understand the algebra on the 2nd step. How does 15^x + 15^x+1 get factored out into 15^x(1+15). Can you explain?

Cheers,
ryguy
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If x and y are integers and (15^x) + ___________________ = 15 Jul 2008, 13:13
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Take an example without variables

4^3 * 4 = 4^4 {4*4*4*4}....or you could write it out before the addition...4^3 * 4^1...add the exponents. So 15^x * 15 = 15^(x+1)
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If x and y are integers and (15^x) + ___________________ = 15 Jul 2008, 14:39
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ryguy904
You are correct. For some reason I don't understand the algebra on the 2nd step. How does 15^x + 15^x+1 get factored out into 15^x(1+15). Can you explain?

Cheers,
ryguy
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This is a factorisation by \(15^x\):

\(15^x + 15^{x+1} = 1 * 15^x + 15 * 15^x = 15^x \left( 1 + 15 \right) = 15^x * 16 = 15^x * 4^2\) ;)
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If x and y are integers and (15^x) + ___________________ = 15 Jul 2008, 17:26
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Oski
You don't know that y=0.

Just simplify the equation:

\(\frac{15^x + 15^{x+1}}{4^y} = 15^y\)

\(15^x \left( \frac{1 + 15}{4^y} \right) = 15^y\)

\(15^x \left( \frac{4^2}{4^y} \right) = 15^y\)

\(15^x \left( \frac{1}{4^{y-2}} \right) = 15^y\)

Now, stem says that x and y are integer. This last equation can holds true only if \(y=2\) in order to get rid of the power of 4 (other than that you just have powers of 3 and 5 : you cannot have a power of 4 on the left or the right of the equation).

Then if \(y=2\) ==> \(x=2\) ==> Answer is (A)
Show more

ah... missed "integer" in the question prompt :roof
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If x and y are integers and (15^x) + ___________________ = 15 Jul 2008, 17:44
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jallenmorris
Take an example without variables

4^3 * 4 = 4^4 {4*4*4*4}....or you could write it out before the addition...4^3 * 4^1...add the exponents. So 15^x * 15 = 15^(x+1)
Show more

This helped. For some reason it didn't click that 15 was the base and x was just some unknown power. So factoring out 15 means you can do 15^x+1.

Bravo!
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If x and y are integers and (15^x) + ___________________ = 15 Jul 2008, 18:45
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i was able to factor to the very end, but if y=2, why does it mean that x=2 ?
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If x and y are integers and (15^x) + ___________________ = 15 Jul 2008, 22:33
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[15^x ][15+1]=[15^y][4^y]
[15^x ][4^2]=[15^y][4^y]
x=y,2=y
->x=2
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If x and y are integers and (15^x) + ___________________ = 15 Jul 2008, 22:36
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Cannot be determined is my pick
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Oski
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If x and y are integers and (15^x) + ___________________ = 15 Jul 2008, 23:36
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pmenon
i was able to factor to the very end, but if y=2, why does it mean that x=2 ?
Show more
From \(15^x \left( \frac{1}{4^{y-2}} \right) = 15^y\), if you set y=2 you are then left with:

\(15^x = 15^2\) ==> \(x=2\) ;)
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If x and y are integers and (15^x) + ___________________ = 16 Jul 2008, 05:22
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Oski
You don't know that y=0.

Just simplify the equation:

\(\frac{15^x + 15^{x+1}}{4^y} = 15^y\)

\(15^x \left( \frac{1 + 15}{4^y} \right) = 15^y\)

\(15^x \left( \frac{4^2}{4^y} \right) = 15^y\)

\(15^x \left( \frac{1}{4^{y-2}} \right) = 15^y\)

Now, stem says that x and y are integer. This last equation can holds true only if \(y=2\) in order to get rid of the power of 4 (other than that you just have powers of 3 and 5 : you cannot have a power of 4 on the left or the right of the equation).

Then if \(y=2\) ==> \(x=2\) ==> Answer is (A)
Show more
Now, stem says that x and y are integer. This last equation can holds true only if in order to get rid of the power of 4 (other than that you just have powers of 3 and 5 : you cannot have a power of 4 on the left or the right of the equation).

Then if ==> ==> Answer is (A)
this is a very good solution. I stopped with the last equation and didn't analyze further seeing the two variables present.



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