# Unknown indice using logarithms

• Nov 22nd 2008, 05:26 AM
BG5965
Unknown indice using logarithms

http://i301.photobucket.com/albums/nn74/BL5965/3d.png

Thanks - solve in terms of x.
• Nov 22nd 2008, 05:34 AM
earboth
Quote:

Originally Posted by BG5965

http://i301.photobucket.com/albums/nn74/BL5965/3d.png

Thanks - solve in terms of x.

$\log_2\left(2^{2x}-56\right) - x=0~\implies~\log_2\left(2^{2x}-56\right) = x$

Now use the base 2 with both sides of the equation:

$2^{2x}-56 = 2^x$

Use substitution $2^x = y$ that means y > 0.

You'll get:

$y^2-y-56 = 0~\implies~y = -7~\vee~y=8$

You only can use the second solution:

$8=2^x~\implies~x=3$