Need help in how to approach exponential equations

Question

How do I approach solving these types of equations

$10^x -5^{x-1}×2^{x-2}=950$

Answer 1

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 $$\left( \begin{array}{cc} n & x_n \\ 0 & 3.54988 \\ 1 & 3.28253 \\ 2 & 3.26147 \\ 3 & 3.26122 \end{array} \right)$$

$$\left( \begin{array}{cc} n & x_n \\ 0 & 4.33467 \\ 1 & 4.26127 \\ 2 & 4.24217 \\ 3 & 4.24102 \end{array} \right)$$

Answer 2

Notice that $10^x$ can be written as $(5\times2)^x=5^x\times2^x$.

Also notice that $5^{x-1}=5^x\times5^{-1}$

Answer 3

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 $x=3.$