Solve $a^x = bx$ without graphing? 
Solve $a^x = bx$ without graphing?

First of all, I am only in 8th grade, and while I do study topics from a much higher level, I do not yet understand calculus. I also have searched various internet resources, but the only solution to this problem seems to be graphing.
I would like to know if a problem in the format of $a^x = bx$ (where $a$ and $b$ are constants) can be solved with a moderate accuracy (such as first 3 decimal digits) without graphing OR trial and error. So far, I could make this problem a logarithmic one: $log_a (bx) = x$, but I am unsure of how to progress from here.
If you need an example, here is one:
$10^x = 7x.$
Furthermore, if the previous is solvable, will it be simple to solve this problem:
$100(1.05)^x = 10x + 50.$
Thanks for your effort and time.

As said in comments, the only explicit solutions of the zero's of function
$$f(x)=a^x-bx$$ are given in terms of Lambert function (which is not elementary) (just have a look here)
$$x=-\frac{1}{\log (a)}W\left(-\frac{\log (a)}{b}\right)$$ and, in the linked page, you will find expansion allowing the evaluation of $W(t)$.
If, for the time being, you cannot use Lambert function, without graphing, you can do good work. For the next, I shall assume that $a$ and $b$ are real and positive.
We have
$$f'(x)=a^x \log (a)-b \qquad \text{and} \qquad f''(x)=a^x \log ^2(a)$$
The first derivative cancels at
$$x_*=\frac{1}{\log (a)}\log \left(\frac{b}{\log (a)}\right)$$
Since $f(0)=1$, if
$$f(x_*)=\frac{b}{\log (a)}\left(1-\log \left(\frac{b}{\log (a)}\right)\right)$$ is negative, then you conceive that there two roots, one on each side of $x_*$. They correspond to the two branches of Lambert function. If this is not the case, no real solution.
For illustration, considering your case $a=10$ and $b=7$, we have $x_*=0.482882$ and $f(x_*)=-0.340115$; then two roots.
If, as in this case, $a$ and $b$ are "similar", you can approximate the solutions expanding $f(x)$ as a Taylor series around $x_*$ and have
$$x_\pm=x_* \pm \sqrt{-2 \frac{f(x_*)} {f''(x_*)} }$$
For the working case, this would give (as estimates)
$$x_-=0.277449 \qquad \text{and} \qquad x_+=0.688316$$ while the exact solutions are respectively be $x_-=0.259894$ and $x_+=0.673319$ which is not too bad (this was done with my phone). Now, if you want to polish the roots, you need a numerical method such as Newton. I give you below the iterates for each root
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 0.277449 \\
 1 & 0.259315 \\
 2 & 0.259893 \\
 3 & 0.259894
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 0.688316 \\
 1 & 0.673998 \\
 2 & 0.673320 \\
 3 & 0.673319
\end{array}
\right)$$
