Question 913131
{{{16^(x) - 4^(x-1) - 3 = 0}}}


{{{(4^2)^x - 4^(x-1) - 3 = 0}}}


{{{4^(2x) - 4^(x-1) - 3 = 0}}}


{{{(4^x)^2 - 4^(x-1) - 3 = 0}}}


{{{(4^x)^2 - (4^x)*(4^(-1)) - 3 = 0}}}


{{{(4^x)^2 - (4^x)*(1/4) - 3 = 0}}}


{{{(4^x)^2 - (1/4)(4^x) - 3 = 0}}}


{{{z^2 - (1/4)z - 3 = 0}}} Let {{{z = 4^x}}}


{{{4z^2 - z - 12 = 0}}} Multiply both sides by the LCD 4


Use the quadratic formula to get {{{z = (1+sqrt(193))/8}}} or {{{z = (1-sqrt(193))/8}}}


The steps for the quadratic formula are lengthy and I feel they would clutter up this solution page. Let me know if you need to see the steps to getting those solutions for z.


-----------------------------------------------------------------------------------


The solutions in terms of z are {{{z = (1+sqrt(193))/8}}} or {{{z = (1-sqrt(193))/8}}}


We ultimately want solutions in terms of x.


Plug in the first solution for z to get the first solution for x.


{{{z = 4^x}}}


{{{(1+sqrt(193))/8 = 4^x}}}


{{{4^x = (1+sqrt(193))/8}}}


{{{x = log(4,((1+sqrt(193))/8))}}}


So the first solution is {{{x = log(4,((1+sqrt(193))/8))}}}


----------------------------------------


Plug in the second solution for z to get the second solution for x.


{{{z = 4^x}}}


{{{(1-sqrt(193))/8 = 4^x}}}


{{{4^x = (1-sqrt(193))/8}}}


{{{x = log(4,((1-sqrt(193))/8))}}}


So the second solution is {{{x = log(4,((1-sqrt(193))/8))}}}


The exact solutions for x are: {{{x = log(4,((1+sqrt(193))/8))}}} or {{{x = log(4,((1-sqrt(193))/8))}}}


Use a calculator to get these approximate solutions {{{x = 0.44825431399819}}} or {{{x = UND}}}


the UND means "undefined" so it turns out that the second solution {{{x = log(4,((1-sqrt(193))/8))}}} isn't really a solution at all. The argument isn't in the domain of log(x)


So let's revise our answer. The only exact solution is {{{x = log(4,((1+sqrt(193))/8))}}}


The only approximate solution is {{{x = 0.44825431399819}}}



Let me know if you need more help or if you need me to explain a step in more detail.
Feel free to email me at <a href="mailto:jim_thompson5910@hotmail.com?Subject=I%20Need%20Algebra%20Help">jim_thompson5910@hotmail.com</a>
or you can visit my website here: <a href="http://www.freewebs.com/jimthompson5910/home.html">http://www.freewebs.com/jimthompson5910/home.html</a>


Thanks,


Jim