Question

(2^x)+(2^-x)=3 Solve for X

Answer 1

Whee
remember
$(x^m)(x^n)=x^{m+n}$

multiply both sides by 2^x
$(2^{2x})+(2^0)=3(2^x)$
$2^{2x}+1=3(2^x)$
minus 3(2)^x from both sides
$2^{2x}-3(2^x)+1=0$
we can use the quadratic formula
use u subsitution where u=2ˣ
$1(2^x)^2-3(2^x)+1=0$
$1(u)^2-3(u)+1=0$
$u^2-3u+1=0$
use quadratic formula
$u=\frac{3+\sqrt{5}}{2}$ and $u=\frac{3-\sqrt{5}}{2}$

solve
subsitte and solve
u=2ˣ

$2^x=\frac{3+\sqrt{5}}{2}$
take ln of both sides
$x(ln(2))=ln(\frac{3+\sqrt{5}}{2})$
divide both sides by ln(2)
$x=\frac{ln(\frac{3+\sqrt{5}}{2})}{ln(2)}$

for other one
$2^x=\frac{3-\sqrt{5}}{2}$
take ln of both sides
$x(ln(2))=ln(\frac{3-\sqrt{5}}{2})$
divide both sides by ln(2)
$x=\frac{ln(\frac{3-\sqrt{5}}{2})}{ln(2)}$



so $x=\frac{ln(\frac{3+\sqrt{5}}{2})}{ln(2)}$ and $x=\frac{ln(\frac{3-\sqrt{5}}{2})}{ln(2)}$
those are the 2 solutions
aprox x=1.38848 and x=-1.338848