Question

Funçao exponencial. Alguém sabe resolver essa equação? 4^x - 3^(x-1/2) = 3^(x+1/2) - 2^(2x-1)
PS.: "^" significa que está elevado.

Answer 1

$4^x - 3^{x-\frac 1 2} = 3^{x+\frac 1 2} - 2^{2x-1}\\\\ 4^x - 3^x\cdot3^{-\frac 1 2} =3^x\cdot 3^{\frac 1 2} - 2^{2x}\cdot2^{-1}\\\\ 4^{x} - 3^x\cdot\dfrac{1}{3^{\frac 1 2}} =3^x\cdot 3^{\frac 1 2} - (2^2)^x\cdot\dfrac{1}{2}\\\\ 4^x - 3^x\cdot\dfrac{1}{\sqrt3} =3^x\cdot \sqrt3 - 4^{x}\cdot\dfrac{1}{2}$

Multiplicando os dois lados por $2\sqrt3$:

$4^x - 3^x\cdot\dfrac{1}{\sqrt3} =3^x\cdot \sqrt3 - 4^x\cdot\dfrac{1}{2}\\\\ 2\sqrt3\cdot4^x - 2\cdot3^x=6\cdot3^x-\sqrt3\cdot 4^x\\\\ 3\sqrt3\cdot4^x =8\cdot3^x$

Vamos dividir toda a equação por $3^x$:

$3\sqrt3\cdot4^x =8\cdot3^x\\\\ 3\sqrt3\dfrac{4^x}{3^x}=8\\\\ 3\sqrt3\left(\dfrac{4}{3}\right)^x=8\\\\ \left(\dfrac{4}{3}\right)^x=\dfrac{8}{3\sqrt3}=\dfrac{4\cdot2}{3\cdot3^{\frac 1 2}}=\dfrac{4\cdot4^{\frac{1}{2}}}{3\cdot3^{\frac1 2}}\\\\ \left(\dfrac{4}{3}\right)^x=\dfrac{4^{1+\frac{1}{2}}}{3^{1+\frac1 2}}\\\\ \left(\dfrac{4}{3}\right)^x=\dfrac{4^{\frac{3}{2}}}{3^{\frac3 2}}\\\\ \left(\dfrac{4}{3}\right)^x=\left(\dfrac{4}{3}\right)^{\frac3 2}$

Como as bases são iguais, podemos igualar os expoentes:

$\boxed{x=\dfrac{3}{2}}$