Solving an algebraic equation for x

Question

$(($ 3^$2\sqrt{3x})$/4$)$ $+3=$ 3^$\sqrt{3x}$

= $($ (3^${2}*{3x^{1/2}}$)/4$)$ $+3=$ 3^${3x^{1/2}}$

After simplifying:

= ($3^{6x^2}$ $+ 3$)/4 $= 3^{3x}$

= $3^{6x} + 3 = 12^{3x}$

I tried solving this problem but I am getting stuck everytime. I tried teaching some of these concepts to myself. I am exhausted and really need help. Can someone please help me with this solution?

Answer 1

$$ is- it ? ,\frac{3^{2*\sqrt{3x}}+3}{4}=3^{\sqrt{3x}}\\if\\yes\\a=3^{\sqrt{3x}}\\then\\\frac{a^2+3}{4}=a\\a^2+3=4a\\a^2-4a+3=0\\(a-1)(a-3)=0\\a=1,a=3\\a=1 ,3^{\sqrt{3x}}=1\\\sqrt{3x}=0 ,x=0\\a=3 ,3^{\sqrt{3x}}=3 \\\sqrt{3x}=1 ,x=\frac{1}{3} $$

Answer 2

If I interpreted your equation correctly, i.e.

$ \frac{1}{4}{3}^{2\,\sqrt {3}\sqrt {x}}+3-{3}^{\sqrt {3}\sqrt {x}}=0 $

Your simplification is incorrect.

After subsitution of \sqrt(3x)=c, solving for c and resubstitution one obtains:

$1/3\,{\frac { \left( \ln \left( 2 \right) +\ln \left( 1-i\sqrt {2} \right) \right) ^{2}}{ \left( \ln \left( 3 \right) \right) ^{2}}}$

and

$1/3\,{\frac { \left( \ln \left( 2 \right) +\ln \left( 1+i\sqrt {2} \right) \right) ^{2}}{ \left( \ln \left( 3 \right) \right) ^{2}}}$

or as a floating point approximation

0.1742853106- 0.6556125383i

and

0.1742853106+ 0.6556125383i