# What are the roots of 4^x + 12*16^x = 11

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Given the equation:

4^x + 12*16^x = 11

To solve the exponent equation we will use the substitution method.

Let us assume that y = 4^x

==> y + 12y^2 = 11

Now we will rewrite with all terms on the left side.

==> 12y^2 + y - 11 = 0

Now we will use the formula to solve for y.

==> y1= [ -1+ sqrt(1 + 4*12*11) / 2*12

= (-1 + sqrt529) / 24

= ( -1 + 23) / 24 = -/24 = 11/12

==> y1= 11/12 ==> -11/12 = 4^x ==> x = log 4 ( 11/12).

==> y2= ( -1 - 23)/24 = -24/24 = -1 ( not a valid solution).

Then the only solution is :

**x = log 4 (11/12) = log (11/12) / log 4 = (log 11 - log 12)/ log 4**

We have to find the roots of 4^x + 12*16^x = 11.

We can write 4^x as y. We know that 16^x = 4^x^2 = y^2

4^x + 12*16^x = 11

=> 12y^2 + y = 11

=> 12y^2 + 12y – 11y – 11 =0

=> 12y (y + 1) – 11 (y + 1) =0

=> (12y – 11) (y + 1) = 0

y = 11/12 and -1.

y = 4^x, we can eliminate y = -1 as 4^x cannot be negative

So we have only 4^x = 11/ 12

Now take the logarithm on both the sides.

x log 4 = log (11 /12)

=> x = log (11/ 12)/ log 4.

**Therefore x is equal to log (11/12) / log 4.**

4^x+12*16^x=11.

To solve this we substitute 4^x = y and 16^x = (4^x)^2 = y^2 in the given equation:t+12t^2= 11.

=> 12t^2+t -11 = 0.

=> (12t-11))(t+1) = 0.

=> 12t-11 = 0. Or t+1 = 0.

=> 12t = 11. Ot t = -1.

12t=11 => t = -1 We ignore t = -1. Or 4^x= -1, as it doesnot give real roots .

So 4^x = 11/12.

We take logarithms:

xlog4 = log(11/12).

x = {**log(11/12)}/log4**