How do you solve a system of equation with a variable as an exponent? So, I've been doing some random math on my own and I ran into a problem. I was met with the system of equations of $$x=3^y$$ $$x+1=5^y$$ How can I solve this algebraically without using the graphs?
 A: In general it's hard to solve these kinds of systems where there are different bases (in your case, $3$ and $5$). In fact, there is no exact solution to this system, but Wolfram Alpha gives the approximate solutions as $x = 2.22302$ and $y = 0.72716$. If the $x + 1$ in the second equation was just an $x$, then we could have been able to solve it using logs, however.
A: So you need $3^y + 1 = 5^y \iff f(y) = \left(\frac35 \right)^y+\left(\frac15 \right)^y = 1$.  Now $f(0) = 2, f(1) < 1 \implies$ as $f$ is continuous we must have at least one solution for  $y \in (0, 1)$. Further as $f$ is strictly decreasing everywhere, there can be only one solution.  
Finding this unique real number $ \approx 0.7272$ remains however a numerical task.      
A: Eliminating $x$ from the first equation and replacing in the second let you with the equation $$3^y+1=5^y$$ which is highly nonlinear (and you look for the intersection of these two functions).
You can make the problem nicer taking logarithms and get $$y\log(5)=\log(3^y+1)$$ which is now the intersection of a curve and a straight line. 
For solving the equation $$f(y)=y\log(5)-\log(3^y+1)=0$$ let us use Newton method which, starting from a "reasonable" guess $y_0$, will update it according to $$y_{n+1}=y_n-\frac{f(y_n)}{f'(y_n)}$$ $$f'(y)=\log (5)-\frac{3^y \log (3)}{3^y+1}$$ By inspection, you know that the solution for $y$ is between $0$ and $1$  since $f(0)=-\log (2)<0$ and $f(1)=\log \left(\frac{5}{4}\right)>0$. So, let us start iterating at $y_0=1$; the successive iterates of Newton method are then $0.7159139357$, $0.7271409790$, $0.7271601514$ which is the solution for ten significant figures.
Now, compute $x$ from $x=3^y$ and get $x=2.223020992$
