Logarithm Problem

**runrun** · June 2nd 2012, 09:45 AM

Hi,

Please could anyone help with this problem please? I know the answers from an answer sheet (x=1 and x=ln12/ln4) , but I have no idea how the answers were reached.

Many thanks

4^2x + 48 = 4^(x+2)

**skeeter** · June 2nd 2012, 09:50 AM

Originally Posted by runrun

Hi,

Please could anyone help with this problem please? I know the answers from an answer sheet (x=1 and x=ln12/ln4) , but I have no idea how the answers were reached.

Many thanks

4^2x + 48 = 4^(x+2)

$4^{2x} + 48 = 4^{x+2}$

$4^{2x} - 4^{x+2} + 48 = 0$

$4^{2x} - 4^2 \cdot 4^x + 48 = 0$

$(4^x)^2 - 16 \cdot 4^x + 48 = 0$

let $u = 4^x$ ...

$u^2 - 16x + 48 = 0$

solve the quadratic for u by factoring ... then solve for x

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