solve equations: 2^x + 2^-x = 5/3 ; 3^(2x+1) - 10*(3^x) +3 = 0?

Question

can u help me solve these two. i know how to solve it... can u do it without a calculator and show me the steps though.

Answer 1

1) 2^x + 2^-x = 5/3 (Eq. 1)
Let's call t = 2^x. Obviously t must >0.
-> 2^-x = 1/t

Substitute this into (Eq. 1), you have t + 1/t = 5/3

<-> (t^2 + 1) / t = 5/3, Now multiply both sides with 3t
<-> 3(t^2 + 1) = 5.t
<-> 3t^2 - 5t + 3 = 0

This equation doesn't have any real solutions. In fact, LHS = (t + 1/t) is always >= 2 (AM-GM inequality) > RHS = 5/3.

2) 3^(2x + 1) - 10 * 3^x + 3 = 0 (Eq. 2)

Call t = 3^x. t>0 3^(2x+1) = 3 * 3^(2x) = 3 * (3^x)^2 = 3 * t^2

Substitute t into (Eq. 2)

3t^2 - 10 t + 3 = 0
<-> t = 3 or t = 1/3
<-> x = 1 or x = -1

Answer 2

It is a perfect square trinomial actually. You will learn to recognize these the more you practice completing the square. x^2 - 2x + 1 = 0 (x - 1)^2 = 0 From here, take the square root of both sides. The square root of 0 is plus or minus 0, but +0 and -0 are the same, so x - 1 = 0 x = 1

Answer 3

Replace 2^x=y. Then first equation reduces to y + 1/y = 5/3 which writes as y^2 - 5y/3 +1 =0. The roots are y1 = (5 + i sqrt(11))/6 and y2 = (5 - i sqrt(11))/6. Terefore, x1 = log2((5 + i sqrt(11))/6), and x2 = log2((5 - i sqrt(11))/6). (I think something is wrong with your coefficients.)

In the second equation we replace 3^x = y and the equation reduces to 3 y^2 - 10 y + 3 =0 giving solutions y1=3 therefore x1 = 1, and y2 = 1/3 therefore x2 = -1.