A logarithm equation

$\\ \log _{ 2 } (2^{ x-1 }+3^{ x+1 }) + \log _{ 2 } (3^{ x }) = 2x$

Now let $n={ 2 }^{ x }$ and let $m={ 3 }^{ x }$ .

Using Properties of logs, we have that

$mn + { 6m }^{ 2 } = 2{ n }^{ 2 }.$

Now divide By mn and let $\cfrac { m }{ n } = t$ .

$1 + \cfrac { 6m }{ n } = \cfrac { 2n }{ m } \\ \\ \cfrac { m }{ n } = t\\ \\ { 6 }t^{ 2 } + t - 2 = 0\\ \\ t = \cfrac { 1 }{ 2 } \quad or \quad t =-\frac{2}{ 3} \quad (\text{rejected because } t \text{ must be positive} )\\ \\ \Rightarrow \boxed { t = \cfrac { 1 }{ 2 } } \\ \\ x = \cfrac { \log { \cfrac { 1 }{ 2 } } }{ \log { \cfrac { 3 }{ 2 } } } = \log _{ 3/2 }{ (1/2) } \\ \\ a + b =\cfrac { 1 }{ 2 } +\cfrac { 3 }{ 2 } = 2$ .

$\log_{2}(2^{x-1}+3^{x+1}) = \log_{2}(2^{2x})-\log_{2}(3^{x}) = \log_{2}(\frac{2^{2x}}{3^{x}})$

$2^{x-1}+3^{x+1} = \frac{2^{2x}}{3^{x}}$

$\frac{2^{x-1}}{3^{x}} + 3 = \frac{2^{2x}}{3^{2x}} = (\frac{2}{3})^{2x}$

$\frac{1}{2}(\frac{2}{3})^{x} + 3 = (\frac{2}{3})^{2x}$

$2(\frac{2}{3})^{2x}-(\frac{2}{3})^{x}-6=0$

$(2(\frac{2}{3})^{x}+3)((\frac{2}{3})^{x}-2) = 0$

$(\frac{2}{3})^{x} = 2, -\frac{3}{2}$

Since $a^{x} > 0 \quad \forall x \in \mathbb{R}, \quad a > 0$ , reject $(-\frac{3}{2})\\$

$\therefore (\frac{2}{3})^{x} = 2$

$x = \log_{\frac{2}{3}}2\\$

Observe that $\log_{m}{n} = \frac{\log_{\frac{1}{m}}n}{\log_{\frac{1}{m}}m} = \frac{\log_{\frac{1}{m}}n}{-1} = -\log_{\frac{1}{m}}n = \log_{\frac{1}{m}}\frac{1}{n}$ . Then

$\boxed{x = \log_{\frac{3}{2}}\frac{1}{2}}$

$a + b = \frac{3}{2} + \frac{1}{2} = \boxed{2}$

Very neat and clear. But I stumbled over the last step. I would find it clearer if you inserted the line

$\frac{1}{2} = ( \frac{3}{2} )^{x}$

$\log_{2}{2^{x-1}+3^{x+1}}=2x-\log_{2}{3^x}$

Let $m=2^x$ and $n=3^x$

Simplifying the above expression we get:

$\dfrac{m}{2}+3n=\dfrac{m^2}{n} \Rightarrow (3n+2m)(2n-m)=0$

$2n=m \Rightarrow 2\times 3^x=2^x \Rightarrow x=\log_{\dfrac{3}{2}}{\dfrac{1}{2}}$

$a=\dfrac{3}{2}, b=\dfrac{1}{2}$

$a+b=\dfrac{3}{2}+\dfrac{1}{2}=2$

$\log_{2}(2^{x-1} + 3^{x+1}) = 2x - \log_{2}(3^x)$

$log_{2}(2^{x-1} + 3^{x+1}) = \frac{2^{2x}}{3^x}$

$2^{x-1}+3^{x+1}$ = $\frac{2^{2x}}{3^x}$

$2^x(\frac{1}{2}$ + 3 $\frac{3^x}{2^x}$ ) = $\frac{2^{2x}}{3^x}$

$\frac{1}{2}$ + 3 $\frac{3^x}{2^x}$ = $\frac{2^{x}}{3^x}$

$\frac{1}{2}$ + 3t = $\frac{1}{t}$

$\frac{6t^2 + t - 2}{2t}$ = 0

t = $\frac{1}{2}$

${\frac{3}{2}}^x$ = $\frac{1}{2}$

x = $\log_{\frac{3}{2}}(\frac{1}{2}$ ))