Question

(1/2)^4x+1=8^2x+1

Answer 1

\[\LARGE \left( \frac{ 1 }{ 2 }\right) ^{4x}+1=8^{2x}+1\] yes?

Answer 2

no, \[(\frac{ 1 }{ 2 })^{4x+1} = 8^{2x+1}\]

Answer 3

Okay, thats why parentheses are important :P

1/2 = 2^-1 and 8 = 2^3, so...

\[\large (2^{-1})^{4x+1} = ({2^3})^{2x+1}\]
Now use this fact: \[\LARGE (x^a)^b = x^{a b}\]

Answer 4

Ohhhh, okay...thank you!

Answer 5

No prob. Did you get -(4x+1) = 3(2x+1) ?

Answer 6

yup