Solving $5000 = 2000(1 + (0.0225/12))^{12t} + 2000(1 + (0.03/12))^{12t}$ So I recently had a math problem in my algebra 2 class:

There are two accounts with a principle of $2000$, compounded monthly, one growing at $3\%$ and one growing at $2.25\%$. Find when the sum of both accounts is $5000$.

I came up with a relatively simple equation: $$5000 = 2000(1 + (0.0225/12))^{12t} + 2000(1 + (0.03/12))^{12t}.$$
When I went to use Logarithms to solve this, I became frustrated upon not finding a single way to isolate both $t'$s. I asked several of my friends in higher math classes, and none of them had a clue. I did some research and after simplifying my equation to $$\log_{(1 + (0.0225/12))}{\left(\frac{5} {2} - (1 + (0.03/12))^{12t}\right)} = 12t$$
I was able to find the identity $$\log{(a + b)} = \log{a} + \log\left({1 + \frac{b}{a}}\right).$$
Which would be able to transform the left equation into something which again has two terms, although one of those terms would be a $1$. I'm not even sure this would be particularly helpful, but it was all I could find in my research. I know the answer to $t$ is $\approx 8.49$, although this was only done with a graphing calculator. I tried Wolfram Alpha, but it didn't even show the answer to the equation.
So far I've talked to two teachers, my algebra 2 teacher, and the BC calculus teacher (who I talk to a lot), and both of them told me that it is only possible through graphing and seeing where the constant equation and the equation of the addition of the exponential formulas intersect. To me this doesn't make any sense though, the solution is available and doesn't look like it would require any iteration like some other problems. Is it true that it is simply impossible to solve this algebraically, or is there some way to solve this?

As already said in comments, you will need some numerical method but you can easily find the range in which the solution is since if 
$$f= 2000(1 + (0.0225/12))^{12t} + 2000(1 + (0.0300/12))^{12t}$$
$$g= 2000(1 + (0.0225/12))^{12t} + 2000(1 + (0.0225/12))^{12t}=4000(1 + (0.0225/12))^{12t}$$
$$h= 2000(1 + (0.0300/12))^{12t} + 2000(1 + (0.0300/12))^{12t}=4000(1 + (0.0300/12))^{12t}$$ you have
$$g < f < h$$ Solving $5000=g$ and $5000=h$ does not make any problem and leads to
$$7.44741 < t <9.92679$$ So, select for $t_0$ some value between these bounds (the average for example, and start Newton method which would converge quite fast as shown below
$$\left(
\begin{array}{cc}
 n & t_n \\
 0 & 8.687099035 \\
 1 & 8.491460917 \\
 2 & 8.490944850 \\
 3 & 8.490944847
\end{array}
\right)$$ In practice, you do not need such accuracy and, more than likely, one or two iterations would be sufficient.
