1. ## logarithms

find x if 3^(log base (a) x)+3*x^(log base(a) 3)=2

2. oh i got the answer we know by property that the two logs in question would be equal

then we can solve it

3. Hello, prasum!

I think I solved it . . . and got an ugly answer.

$\displaystyle \text{Find }x\text{ if: }\;3^{\log_ax} +3\cdot x^{\log_a3} \:=\:2$

$\displaystyle \text{Let: }\:3^{\log_ax} \:=\;P$

$\displaystyle \text{Take logs, base 3: }\;\log_3\!\left(3^{\log_ax}\right) \;=\;\log_3(P)$

. . . . . . . . . . . $\displaystyle \log_a(x)\cdot\underbrace{\log_3(3)}_{\text{This is 1}} \;=\;\log_3(P)$
. . . . . . . . . . . . . . . . . $\displaystyle \log_a(x) \;=\;\dfrac{\log_a(P)}{\log_a(3)}$

. . . . . . . . . . . . . . . . .$\displaystyle \log_a(P) \;=\;\log_a(x)\cdot\log_a(3)$

. . . . . . . . . . . . . . . . . . . . .$\displaystyle P \;=\;a^{\log_a(x)\cdot\log_a(3)}$

. . . . . . . . . . . . . . . . . . . . .$\displaystyle P \;=\;\left(a^{\log_a(x)}\right)^{\log_a(3)}$

. . . . . . . . . . . . . . . . . . . . .$\displaystyle P \;=\;x^{\log_a(3)}$

Hence: .$\displaystyle 3^{\log_a(x)} \;=\;x^{\log_a(3)}$

The equation becomes: .$\displaystyle x^{\log_a(3)} + 3\cdot x^{\log_a(3)} \;=\;2$

. . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle 4\cdot x^{\log_a(3)} \;=\;2$

. . . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle x^{\log_a(3)} \;=\;\dfrac{1}{2}$

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle x \;=\;\left(\dfrac{1}{2}\right)^{\frac{1}{\log_a(3) }}$

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle x \;=\;\left(\dfrac{1}{2}\right)^{\log_3(a)}$