In GMAT Review Book there is a DS question as follows: If x is an integer, is 9^x + 9^-x=b?
1) 3^x + 3^-x=square root of (b+2)
2) x>0=b
it says its helpful to know (xr)^2=x^2r just prior to the answer reasoning and was wondering if this is a typo as (xr)^2 is x^2 r^2 but not x ^2r unless i am misunderstanding something.
Then in the solution it says (3^x + 3^-x)^2=b+2 and then the next line says 3^2x +2(3^x * 3^-x) +3^-2x=b+2.....could someone kindly tell me how to get that from the line above it....i am confused and missing something. thanks
DS Question
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I think there is a typo in stmt II since x>0=b doesnt make much sense.
Coming to the problem:
One way of doing it:
Q:
Is 9^x + 9^-x = b
can be rephrased as
Is 3^2x + 1/3^2x = b {Exponent rule : a^-x = 1/ a^x and also (a^b)^c = a^(b*c) where a = 3 b=2 i.e 9 can be written as 3^2 and c = x}
Stmt I
3^x+ 3^-x = sqrt(b+2) {Exponent rule : a^-x = 1/ a^x }
3^x + 1/3^x = sqrt(b+2)
Square both sides
3^2x + 1/3^2x + 2*3^x * 1/3^x = b+2 {(a+b) ^ 2 = a^2+b^2+2ab where a = 3^2x b = 1/3^2x }
3^2x + 1/3^2x + 2 = b+2
Subtract 2 from both sides
3^2x + 1/3^2x = b
SUFF since we rephrased the question
Other way is to to get stmt I to match up to the actual question:
Is 9^x + 9^-x = b
Is 9^x + 1/9^x = b
Stmt I
3^x+ 3^-x = sqrt(b+2)
3^x + 1/3^x = sqrt(b+2) {(a+b) ^ = a^2+b^2+2ab where a = 3^x b = 1/3^x }
Square both sides
3^2x + 1/3^2x + 2*3^x*1/3^x = b+2
3^2x + 1/3^2x + 2 = b+2 {3^2x can be rewritten as 9^x using exponent rule a^(b*c) = (a^b)^c where a=3 b=2 c=x -> Just a rephrase of the rule (a^b)^c = a^(b*c) }
9^x+ 1/9^x = b
SUFF
Hope this helps!
IMO should read (x^r)^2 = x^2rit says its helpful to know (xr)^2=x^2r just prior to the answer reasoning and was wondering if this is a typo as (xr)^2 is x^2 r^2 but not x ^2r unless i am misunderstanding something
Coming to the problem:
One way of doing it:
Q:
Is 9^x + 9^-x = b
can be rephrased as
Is 3^2x + 1/3^2x = b {Exponent rule : a^-x = 1/ a^x and also (a^b)^c = a^(b*c) where a = 3 b=2 i.e 9 can be written as 3^2 and c = x}
Stmt I
3^x+ 3^-x = sqrt(b+2) {Exponent rule : a^-x = 1/ a^x }
3^x + 1/3^x = sqrt(b+2)
Square both sides
3^2x + 1/3^2x + 2*3^x * 1/3^x = b+2 {(a+b) ^ 2 = a^2+b^2+2ab where a = 3^2x b = 1/3^2x }
3^2x + 1/3^2x + 2 = b+2
Subtract 2 from both sides
3^2x + 1/3^2x = b
SUFF since we rephrased the question
Other way is to to get stmt I to match up to the actual question:
Is 9^x + 9^-x = b
Is 9^x + 1/9^x = b
Stmt I
3^x+ 3^-x = sqrt(b+2)
3^x + 1/3^x = sqrt(b+2) {(a+b) ^ = a^2+b^2+2ab where a = 3^x b = 1/3^x }
Square both sides
3^2x + 1/3^2x + 2*3^x*1/3^x = b+2
3^2x + 1/3^2x + 2 = b+2 {3^2x can be rewritten as 9^x using exponent rule a^(b*c) = (a^b)^c where a=3 b=2 c=x -> Just a rephrase of the rule (a^b)^c = a^(b*c) }
9^x+ 1/9^x = b
SUFF
Hope this helps!