MHB -s6.6 solve exponential eq 3^x-14\cdot 3^{-x}=5

[karush]

\tiny{s464\\6.6}
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

 

[HOI]

That's good. I would have done it just slightly differently. I would have let y= 3^x so that the equation became y- 14/y= 5. multiplying by y gives y^2- 5y- 14= 0. That, of course, factors as (y- 7)(y+ 2)= 0 so that y= 7 or y= -2. But 3^x cannot be negative so the only solution is 3^x= 7, x log(3)= log(7) and x= log(7)/log(3) as you had.

(I don't see where you "factored out" a 3.)