Solving $2^x + 3^x = 12$ I need to solve $2^x + 3^x = 12$ for real $x$. I tried the following:
$$
2^x + 3^x = 3\times 2^2 \\
1+(3/2)^x= 3\times 2^{2-x}
$$
But from here on I don't know how to apply logarithms.
 A: As stated in the comments, this equation cannot be solved by algebraic means, but we can use the Newton-Raphson method to get arbitrarily close to the root.
The Newton-Raphson iteration, used to solve equations of the form $f(x)=0$, is as follows: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$
To find the root of your equation, re-arrange it so that the $RHS$ is 0, and then define $f(x)=LHS$. 
$\therefore f(x)= 2^x + 3^x -12$
$\implies f'(x)= ln(2)·2^x + ln(3)·3^x$
A good guess for our first value of $x$ is $2$, as when $x=2, f(x)=1$. I'll aim to get the root to $18$ decimal places.
$\therefore$ Let $x_1=2$.
$\therefore$ Via Newton-Raphson: $x_2 = 2-\frac{2^2 + 3^2 -12}{ln(2)·2^2 + ln(3)·3^2}=1.921011678131129049981...$, which is the value of $x_2$. 
Since I decided to aim for $18$ decimal places, I must apply the algorithm repeatedly until I get the same result to $18$ decimal places in two successive iterations. My values for $x_3$, $x_4$, $x_5$ and $x_6$ are:
$x_3=1.917691510628841444604$
$x_4=1.917685944904149110065$
$x_5=1.917685944888544576369$
$x_6=1.917685944888544576368$
Hence the root of the equation $2^x + 3^x = 12$ is:
$x=1.917685944888544576$
A: Value of $x=1.917685944888545$
(15 digits round off)

