Question

25^x + 35^x = 49^x
Passo a passo por favor.

Answer 1

Resposta:

$\sf 25^x+35^x=49^x$

$\sf \dfrac{25^x}{25^x}+\dfrac{35^x}{25^x}=\dfrac{49^x}{25^x}$

$\sf 1+\bigg(\dfrac{35}{25}\bigg)^{\!\!x}=\bigg(\dfrac{49}{25}\bigg)^{\!\!x}$

$\sf 1+\bigg(\dfrac{7}{5}\bigg)^{\!\!x}=\bigg(\dfrac{7^2}{5^2}\bigg)^{\!\!x}$

$\sf 1+\bigg(\dfrac{7}{5}\bigg)^{\!\!x}=\bigg(\dfrac{7}{5}\bigg)^{\!\!2x}$

Por mudança de variável, aplique o artifício k = (7/5)ˣ:

$\sf 1+k=k^2$

$\sf k^2-k-1=0$

$\sf k=\dfrac{-\,b\pm\sqrt{b^2-4ac}}{2a}$

$\sf k=\dfrac{-\,(-1)\pm\sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

$\sf k=\dfrac{1\pm\sqrt{1+4}}{2}\Rightarrow k=\dfrac{1\pm\sqrt{5}}{2}$

Voltando a (7/5)ˣ = k:

$\sf \bigg(\dfrac{7}{5}\bigg)^{\!\!x}=k$

$\sf \bigg(\dfrac{7}{5}\bigg)^{\!\!x}=\dfrac{1\pm\sqrt{5}}{2}$

Aplique logaritmo na base 7/5 em ambos os membros a fim de usar a propriedade da potência para tirar x do expoente.

$\sf log_{\frac{7}{5}}\bigg(\dfrac{7}{5}\bigg)^{x}=log_{\frac{7}{5}}\bigg(\dfrac{1\pm\sqrt{5}}{2}\bigg)$

$\sf x\cdot log_{\frac{7}{5}}\bigg(\dfrac{7}{5}\bigg)=log_{\frac{7}{5}}\bigg(\dfrac{1\pm\sqrt{5}}{2}\bigg)$

$\sf x\cdot 1=log_{\frac{7}{5}}\bigg(\dfrac{1\pm\sqrt{5}}{2}\bigg)$

$\sf x=log_{\frac{7}{5}}\bigg(\dfrac{1\pm\sqrt{5}}{2}\bigg)$

Resta agora:

$\sf x=log_{\frac{7}{5}}\bigg(\dfrac{1+\sqrt{5}}{2}\bigg) ~ou~x=log_{\frac{7}{5}}\bigg(\dfrac{1-\sqrt{5}}{2}\bigg)$

Dado que não é possível 7/5 elevado a qualquer número real resultar num valor negativo, temos que

$\sf x\neq log_{\frac{7}{5}}\bigg(\dfrac{1-\sqrt{5}}{2}\bigg)}$

$\red{\boxed{\sf x=log_{\frac{7}{5}}\bigg(\dfrac{1+\sqrt{5}}{2}\bigg)}\sf~\rightarrow~Soluc_{\!\!_s}}\tilde{\sf a}\sf o$