Indices past paper question help

• May 5th 2012, 07:42 PM
mathkid12
Indices past paper question help
By using y=2^x or otherwise, solve

4^x - 3 (2^(x+1)) + 8= o

so i understand to convert the 4^x to 2^x(2) but how to proceed ??
sorry cant type equations properly dont know what language this forums use and again sorry.
• May 5th 2012, 07:51 PM
Prove It
Re: Indices past paper question help
Quote:

Originally Posted by mathkid12
By using y=2^x or otherwise, solve

4^x - 3 (2^(x+1)) + 8= o

so i understand to convert the 4^x to 2^x(2) but how to proceed ??
sorry cant type equations properly dont know what language this forums use and again sorry.

\displaystyle \displaystyle \begin{align*} 4^x - 3 \cdot 2^{x + 1} + 8 &= 0 \\ \left( 2^2 \right) ^x - 3 \cdot 2 \cdot 2^x + 8 &= 0 \\ 2^{2x} - 6 \cdot 2^x + 8 &= 0 \\ \left( 2^x \right) ^2 - 6 \cdot 2^x + 8 &= 0 \\ X^2 - 6X + 8 &= 0 \textrm{ after making the substitution } X = 2^x \end{align*}

You should now be able to solve.
• May 5th 2012, 08:05 PM
mathkid12
Re: Indices past paper question help
thanks, thank you very very much!!!!!!!!1