Help with logarithmic equation? Find the value of x if: 
$(2^x)^2 + 3(2^x) - 18 = 0$
So far, I have done 
$(2^x)^2 + 2^x(3)=18$
$(2^x)^2+2^x=6$
What should i do with $(2^x)^2+2^x$ so 
i can  have only one $^X$ on the left side of the equation?

Let $A = 2^x$. Then the equation becomes
$$A^2 + 3A - 18 = 0$$
which is just one of your everyday quadratic equations. Now, solve the quadratic equation for $A$ to find that $A = -6, 3$. Equate both with $2^t$ to find that
$$2^t = -6, 3$$
For the case when $2^t = 3$, we have $2^t = 3 \implies t = \log_2{3}$.
We reject the case for which $2^t = -6$, because $2^t$ is always positive for real $t$.
