Question

(1/8)^x-1 = 2^2^3-2x^2 see equation drawn below

Answer 1

\[\left(\begin{matrix}1 \\ 8\end{matrix}\right)^{x-1} = 2^{3-2x ^{2}}\]

Answer 2

1/8 is a fraction

Answer 3

\[(\frac{1}{8})^{x-1} = 2 ^{3-2x^2}\]\[(\frac{1}{2^3})^{x-1} = 2 ^{3-2x^2}\]\[(2^{-3})^{x-1} = 2 ^{3-2x^2}\]\[-3(x-1) = 3-2x^2\]\[-3x+3 = 3-2x^2\]\[2x^2-3x = 0\] Can you solve the equation now?
After solving the equation, remember to plug in the number and see if the equation is still true

Answer 4

It says to find all real numbers that satisfy the equation. How do I get that? Says the answer is 0, 3/2

Answer 5

nevermind, i got it! :)

Answer 6

\[2x^2-3x = 0\]\[x(2x-3)=0\]\[x=0 \ or \ (2x-3)=0\]\[x=0 \ or \ x=\frac{3}{2}\]

Answer 7

Sorry that I come too late :(