Need help in how to approach exponential equations How do I approach solving these types of equations
$10^x -5^{x-1}×2^{x-2}=950$
 A: Notice that $10^x$ can be written as $(5\times2)^x=5^x\times2^x$.
Also notice that $5^{x-1}=5^x\times5^{-1}$
A: Consider that you look for the zero's of function
$$f(x)=10^x -5^{x-1}×2^{2x-2}-950=10^x-20^{x-1}-950$$ for which
$$f'(x)=10^x \log (10)-20^{x-1} \log (20)\qquad \text{and} \qquad f''(x)=10^x \log ^2(10)-20^{x-1} \log ^2(20)$$
The first derivative cancels at
$$x_*=2+\frac{\log \left(\frac{5 \log (10)}{\log (20)}\right)}{\log (2)}\approx 3.94227 $$ At this point $f(x_*)=1075.8$ and $f''(x)=-13973.8$ which means that $x_*$ corresponds to a maximum and then two solutions such that $x_1 < x_*$ and $x_2 >x_*$.
To start the search, we need estimates. For that, let us use a series axpansion around $x_*$ to get
$$f(x)=f(x_*)+\frac 12 f''(x_*) (x-x_*)^2+O\left((x-x_*)^3\right)$$ and then the estimates
$$x_1=x_*-\sqrt{-2 \frac{f(x_*)} { f''(x_*)}}\qquad \text{and} \qquad x_2=x_*+\sqrt{-2 \frac{f(x_*) } { f''(x_*)}}$$
Thsi would give as estimates $x_1=3.54988$ and $x_2=4.33467$.
Now, we have all elements to start Newton method which will generate the following iterates
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 3.54988 \\
 1 & 3.28253 \\
 2 & 3.26147 \\
 3 & 3.26122
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 4.33467 \\
 1 & 4.26127 \\
 2 & 4.24217 \\
 3 & 4.24102
\end{array}
\right)$$
A: We have that $$5^{x-1}×2^{x-2}=\frac155^x×\frac142^x=\frac{1}{20}(5×2)^x=\frac{10^x}{20}.$$
Thus your equation becomes $$10^x-\frac{10^x}{20}=950,$$ which can be easily solved for $10^x$ to give $$10^x=1000=10^3.$$ Consequently, we see that $x=3.$
