DS Question

I think there is a typo in stmt II since x>0=b doesnt make much sense.

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

IMO should read (x^r)^2 = x^2r


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!