How does one solve the equation $3^x+5\cdot3^x\cdot2^x-2^x=0$?

Question

I've tried to solve equations $$3^x+5\cdot3^x\cdot2^x-2^x=0$$ $$81^x−2·54^x−36^x−2·24^x+16^x=0$$ but I failed.

I don't know where to start. Any help is welcome

Answer 1

The second equation can be solved by dividing every term by $16^x$ and substituting $$u=\left(\frac 32\right)^x$$ Then we have to solve the equation $$u^4-2u^3-u^2-2u+1=0$$ $$\Rightarrow (u^2-3u+1)(u^2+u+1)=0$$

So the real solutions are given by $$u=\frac {3\pm\sqrt{5}}{2}$$ from which you can deduce the values of $x$

Answer 2

I am going to go ahead and finish up Equation 1.

Having identified one solution $x=-2$, prove it is unique among the reals.

Divide the equation by $2^x$ and move the last, now constant term, to the RHS:

$(3/2)^x+5(3^x)=1$

The LHS is strictly monotonic for all real $x$ so there can be only the one solution.

On to Eq. 2. Hint: There are no rational solutions, yet all solutions can be expressed with elementary functions only. Enjoy!