A few math problems (exponential equations)...?

Question

1. 2^(x-1)-2^x=2^(-3) [answer is no solution]
2. If f(x)= 2^(3x+1), prove that f(x+1) - f(x) is divisible by 7 if x is a non-negative integer.

I'm not really asking for answers, please tell me how to approach the question...(for example, a question like 3(3^x)+9(3^-x)=28, hint would be "let y=3^x") but if you can solve the question then please show me how to do it step by step. Oh yes, and no logarithms please >< haven't learned that yet. Thanks a bunch :)

Answer 1

An exponent with a variable+integer value can be written as the product of the exponent to the integer value and the exponent to the variable.

For example 2^(x+2) = 4* 2^x

1) 1/2 * 2^x - 2^x = 1/8
-1/2 * 2^x = 1/8
2^x = -1/4
Oops, 2^x will never be negative for real x.

2) f(x) = 2^(3x+1)
f(x+1) = 2^(3(x+1)+1) = 2^(3x+4) - 8*2^(3x+1)
f(x+1) - f(x) = 8*2^(3x+1) - 2^(3x+1) = 7*2^(3x+1)

Answer 2

1. 2^(x-1)-2^x=2^-3
Factor 2^(x-1) on the left to get
2^(x-1) (1- 2)= 2^-3
That is,
-2^(x-1)=2^-3,
Which is impossible, since the expression on the left
is a negative number while that on the right is a positive one.
Thus, the problem has no solution.

2. f(x+1)-f(x)=2^ (3(x+1)+1)-2^(3x+1) (by definition)
=2^(3x+3+1)-2^(3x+1) (simplify)
=2^(3x+1) (2^3-1) (factor out)
=2^(3x+1) (8-1) (simplify)
=7. 2^(3x+1) (simplify and change order)
Since x is a non-negative integer, 2^(3x+1) is a positive
integer; thus, f(x+1)-f(x) is divisible by 7.