[Ganhad]

2b) 5^3x * 25^x-2 = 1
2^(3x-1) * 8^(x-1)= 128

5) 81^(x)/27^(2x+1) = 9^(2x-5)/729 Solve This .

7) Fermat stated that every whole number could be expressed as the sum of no more thakt four squares , so that, for example
15=3^2+2^2+1^2+1^2
139=9^2+7^2+3^2
327=13^2+10^2+7^2+3^2

do you think he was correct can you find a general formula for those numbers that require the sum of atleast four squares to make up the number itself?

Help

[TenOfThem]

What do you want help with

Proving Fermat?

[Ganhad]

2b) which i carry on getting the wrong answer and 5 i get for 2b = 15/2 and 5) 13/8 7) i dont get it

[TenOfThem]

(Original post by Ganhad)
2b) 5^3x * 25^x-2 = 1
2^(3x-1) * 8^(x-1)= 128

5) 81^(x)/27^(2x+1) = 9^(2x-5)/729

then apply the rules of indices

ditto for the others ... you need to get everything with the same base

[Ganhad]

after that i get it wrong and that doesnt work for the others

[TenOfThem]

What do you do after that

What is the actual answer if x is not 0.8