- #1

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solve the exponential equation. $3^x-14\cdot 3^{-x}=5$

Rewrite $\quad 3^x+\dfrac{14}{3^x}=5$

$\times$ $\quad 3^x \quad 3^{2x}+14=5\cdot 3^x$

quadratic $\quad 3^{2x}-5\cdot3^x-14 =0$

Factor $\quad (3^x-7)(3^x+2)=0$

Discard $\quad 3^x =7$

hence $\quad x=\log_3(7)=\dfrac{\log 7}{\log 3}\approx 1.7712$

ok i think it is correct

probably better to use () rather than $\cdot$

didn't know is factoring out the 3 was possible