$3^{x}\times8^{\frac{x}{x+2}}=6$
- First I applied a logarithm with base 10 to both sides $\log{3^{x}\times8^{\frac{x}{x+2}}}=\log6$
- $x\log3+\frac{x}{x+2}\log8=x\log6+2\log6$
- $x^2\log3+x(2\log2-\log3)-2\log2=0$
- How can I finish the problem without applying the quadratic formula
$3^x8^{\frac x{x+2}}= 3^x 2^{\frac{3x}{x+2}}= 6 =2*3$
so any solution to $x = \frac {3x}{x+2} =1$ or $x =1$ will be a solution. But is it the only one?
Instead of doing base $10$ do base $2$ or $3$.
$\log_3 3^x + \log_3 8^{\frac x{x+2}} = \log_3 2*3$
$x + \frac x{x+2} \log_3 8 = \log_3 2 + \log_3 3$
$x + \frac {3x}{x+2}\log_3 2 = \log_3 2 + 1$
$x -1 +(\frac {3x}{x+2}-1)\log_3 2 = 0$
$x -1 +\frac {3x -x -2}{x+2} \log_3 2=0$
$x-1 +\frac {2x-2}{x+2} \log_3 2=0$
$(x-1)(x+2) + (2x-2)\log_3 2 = 0$
$x^2 + x(1+2\log_3 2) -(2+2\log_3 2) = 0$.
Quadratic formula (why were you trying to avoid it?)
$x = \frac {-1-2\log_3 2 \pm \sqrt{(1+2\log_3 2)^2 + 4*2*(1+\log_3 2)}}2=$
$\frac {-1-2\log_3 2 \pm \sqrt{9 + 12\log_3 2+ 4(\log_3 2)^2 }}2=$
$\frac {-1-2\log_3 2 \pm (3+2\log_3 2)}2=$
$\{\frac {(-1+3)}2; \frac {-4 - 4\log_3 2}2\}$
$= \{1, -2-2\log_3 2\}$.
.....
If we check
$3^18^{\frac 1{1+2}} = 3*2 = 6$.
and $3^{-2-2\log_3 2}8^{\frac {-2-2\log_3 2}{-2\log_3 2}}=$
$\frac 19\frac 14*8^{\frac {1+\log_3 2}{\log_3 2}}=$
$\frac 1{36}8^{1+\frac 1{\log_3 2}} =$
$\frac 8{36}*8^{\frac 1{\log_3 2}}=$
$\frac 29 *8^{\log_2 3}=\frac 29*2^{(\log_2 3)\cdot 3}= \frac 29*3^3=6$
(remember $\frac 1{log_b a} =\log_a b$ because if $\log_a b =k$ then $a^k=b$ and $a = b^{\frac 1k}$ and $\log_b a = \frac 1k$)
Note: $k = -2-2\log_3 2$ may be written in multiple forms depending ony you we wished to solve this.
Had we chosen to use base $10$ or base $3$ we would have gotten different expressions but they would have the same value.
Example $x^2\log3+x(2\log2-\log3)-2\log2=0$ implies
$x = \frac {\log 3 - 2\log 2 \pm \sqrt{(2\log 2-\log 3)^2 +8\log 2\log 3}}{2\log 3}=$
$\frac {\log 3-2\log 2\pm (2\log 2+\log 3)}{2\log 3}=$
$\{1, \frac {-2\log 3-2\log2}{\log 3}$ and $\frac {-2\log 3-2\log2}{\log 3} = -2-2\log_3 2$
(Hmmm.... somewhere we made an arithmetic error al lost a $\log 3$ but....