Is there a standard way to solve $49^x-42^x=-7$ I was teaching my sister indices a while back. Then after covering a few laws of indices she came up with this equation from the top of her head.

$49^x-42^x=-7$

Of course I couldn't solve it. I figured the answer(s) is in $\mathbb{C}$ since $f(x)=49^x-42^x+7$ does not cross the x axis at all.
The best I could do was shrink this down to:
$7^x(7^x-6^x)=-7$
Is there a way to do this or is it outright impossible. I'm also kind of a beginner in complex analysis so I'm curious to know if the answer isn't as complicated as I think it is.
 A: From $7^x(7^x-6^x) = -7$, since $7^x > 0$ for all $x$ and $7^x > 6^x$ for $x>0$, the left-hand side is positive for all $x>0$, so there is no solution when $x > 0$.  $x = 0$ is not a solution.  For $x < 0$, the left-hand side is the product of $7^x \in (0,1)$ and $7^x - 6^x \in (-1,1)$, so can never be as large in magnitude as $\pm 7$.  Thus, there is no solution in the reals.
In the complexes, \begin{align*} 
49^x - 42^x &= 2\mathrm{i} \cdot \frac{1}{2\mathrm{i}} \left(\frac{49^x}{49^{x/2} 42^{x/2}} - \frac{42^x}{49^{x/2} 42^{x/2}}\right) 49^{x/2} 42^{x/2}  \\
  &= 2\mathrm{i} \cdot \frac{1}{2\mathrm{i}} \left(\frac{49^{x/2}}{42^{x/2}} - \frac{42^{x/2}}{49^{x/2}}\right) (49 \cdot 42)^{x/2}  \\
  &= 2\mathrm{i} \sin \left(\frac{-\mathrm{i}}{2}\ln(49/42) x \right)(49 \cdot 42)^{x/2}
\end{align*}

and replacing the original left-hand side with this, it is straightforward to solve for $x$.
Was missing the $x$ in the sine when I left for lunch.  Realized during and corrected now.  So this doesn't work.
OP's request for Midgardians:  One definition for the complex sine is 
$$  \sin(z) = \frac{\mathrm{e}^{\mathrm{i}z} - \mathrm{e}^{-\mathrm{i}z}}{2\mathrm{i}}  \text{.}  $$
Notice that this is the difference of two reciprocals divided by $2\mathrm{i}$.  Then \begin{align*}
\frac{49^{x/2}}{42^{x/2}} &= \exp \left( \ln\left(  \frac{49}{42}  \right)  \right)^{x/2}  \\
    &= \exp\left( \frac{1}{2} \ln(49/42) x\right)  \\
    &= \exp\left(\mathrm{i} \frac{-\mathrm{i}}{2} \ln(49/42) x\right)
\end{align*}
and the other term is the reciprocal of this one.
A: There are in fact infinitely many complex solutions, which appear to all be in the strip $0.3 \le \text{Re}(z) \le 1$.
Here is a plot of some of them:

Here is a proof that there are infinitely many.  Consider the function $$f(z) = 49^z - 42^z = 42^z ((49/42)^z - 1) $$
This is an entire function, and $f(z+ip) = 42^{ip} f(z) \tag{1}$
where $p = 2 \pi/ \ln(49/42)$.  It has an essential singularity at $\infty$.
By the Great Picard Theorem, $f$ takes all complex values, with at most one exception, in any neighbourhood of $\infty$.  But by (1), if $w$ is an exception then so is $42^{ip} w$.  Since there is at most one exception, it can't be any nonzero value (in fact $0$ is not an exception either).
EDIT: By considering $|f(z)|$, it's easy to see the roots are on the curve (with $z = x + i y$)
$$ 42^{2x} + 49^{2x} - 2 (42\cdot 49)^x \cos(y \ln(7/6)) = 49$$
and since $-1 \le \cos(y \ln(7/6)) \le 1$ we find that the upper bound for $x$ is indeed $1$ (as for $x=1$ and $\cos(y \ln(7/6))$ the $0$), while the lower bound is the positive real solution of $49^x + 42^x = 7$, approximately $0.328316268$, which doesn't appear to have a closed form.
A: I mentioned the need for numerical methods: worse still, there's a lot more than one solution in $\Bbb C$, making this harder to explore. For what it's worth, I wrote some Python to seek a solution by Newton-Raphson. (The code prints pairs of values of $x,\,49^x-42^x+7$ for a while; when it stops, the first value is an approximate root and the second value is very small.) It was approximately $0.501692581938923+7.016931344755486j$ (note $j$ is used in Python instead of $i$). However, multiple solutions exist; when I ran again, I got $0.9889567094750864-41.09191927909876j$.
