Question

Sejam f(x) = $3^{r-1}$. g(x) = $3^{r}$, e s(x)= f(x) + g(x)
Qual é o valor de x tal que s(x) = 4?

Me ajudem, eu tava estudando, mas não sei como fazer essa

Answer 1

$f(x)=3^{x-1}\ \ \ \ g(x)=3^x\ \ \ \ s(x)=f(x)+g(x) \ \ \ \ s(x)=4\\ \\s(x)=f(x)+g(x)\\ \\4=3^{x-1}+3^x\\ \\4=\frac{3^x}{3}+3^x\\ \\4=3^x\cdot \left(\frac{1}{3}+1\right)\\ \\4=3^x\cdot \left(\frac{1}{3}+\frac{3}{3}\right)\\ \\4=3^x\cdot \left(\frac{1+3}{3}\right)\\ \\4=3^x\cdot \left(\frac{4}{3}\right)\\ \\3^x=\frac{4\cdot3}{4} \\3^x=3\\ \\x=1$
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$f(1)=3^{1-1}\ \ \ \ g(1)=3^1\ \ \ \ s(1)=f(1)+g(1) \\f(1)=3^0\ \ \ \ g(1)=3\ \ \ \ s(1)=f(1)+g(1) \\f(1)=1\ \ \ \ g(1)=3\ \ \ \ s(1)=1+3 \\f(1)=1\ \ \ \ g(1)=3\ \ \ \ s(1)=4$

Answer 2

Boa tarde Leticia!

Solução!

$f(x)= 3^{x-1}\\\\\\\ g(x)=3^{x} \\\\\\ s(x)=4\\\\\ f(x)+ g(x)=4$

$3^{x-1}+3^{x}=4\\\\ 3^{x}.3^{-1}+3^{x}=4\\\\\ 3^{x}(3^{-1}+1)=4\\\\\\ 3^{x}( \frac{1}{3} +1)=4\\\\\ 3^{x}( \frac{1}{3} +3)=4\\\\\ 3^{x}( \frac{4}{3})=4\\\\\ 3^{x}= \dfrac{4}{ \dfrac{4}{3} }\\\\\\\\ 3^{x}=4. \frac{3}{4}\\\\\ 3^{x}=3\\\\\ \boxed{x=1 }$

Boa tarde!
Bons estudos!