[{ 7^x + 2.(7^x) + 3.(7^x) + 4.(7^x) + ... + 49.(7^x) } / 25 ] = 343 nilai x yang memenuhi persamaan.... Question from @gwenlauren36

October 2018 2 20 Report

[{ 7^x + 2.(7^x) + 3.(7^x) + 4.(7^x) + ... + 49.(7^x) } / 25 ] = 343 nilai x yang memenuhi persamaan adalah ... help please, thnks x

izharulhaq78 $\displaystyle \frac{7^x+2(7^x)+3(7^x)+...+49(7^x)}{25}=343\\\\\text{misal :}\\7^x=a\\\\\frac{a+2a+3a+...+49a}{25}=343\\\frac{\dfrac{49}{2}(a+49a)}{25}=343\\\frac{\dfrac{49}{2}(50a)}{25}=343\\\frac{49(50a)}{2\times25}=343\\\frac{49(50a)}{50}=343\\49(a)=343\\a=\frac{343}{49}\\a=7\\\text{masukkan kembali nilai }a\\7^x=7\\7^x=7^1\\\\\boxed{\boxed{\therefore x=1}}$

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bgimma33 $$\begin{align}\frac{7^{x}+2\bullet7^{x}+3\bullet7^{x}+4\bullet7^{x}+........+49\bullet7^{x}}{25}~~&=~343\\ \\ \bold{MISALKAN~~~7^x~=~a~~maka~menjadi~~:} \\ \\ \frac{a+2a+3a+4a+.........................+49a}{25}~~&=~343 \\ \bold{S_n~barisan~aritmatika~,~dimana~S_n~=~\frac{n}{2}(U_1+U_n)} \\ \bold{a+2a+3a+4a+.........................+49a~~&=~S_{49}} \\ a+2a+3a+4a+.........................+49a~~&=~\frac{49}{2}(a+49a) \\ a+2a+3a+4a+.........................+49a~~&=~1225~a\end{align}$

$$\begin{align}\frac{a+2a+3a+4a+.........................+49a}{25}~&=~\frac{\not1225a~~^{49a}}{\not{25}~~^1}~=~343 \\ \\ \bold{didapat~:} \\ \\ 49a~&=~343 \\ a~&=~\frac{343}{49} \\ a~&=~7\end{align}$

$\boxed{\bold{a~=~7^x~=~7}} \\ \\ \text{Maka~x~=~1}$

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