how to solve equations with exponents

when we solve an equation with exponents for a variable $x$, we see if the exponents of the variable are of the same exponent.
if same, make each side of the equation have the same base
we can also have
if the exponents are 1 and 2, form a quadratic to find solutions
if the exponents are something else, try to make the exponents be 1 and 2, with the help of introducing new variables as substitution.

worked example: solve $3^{-4x}=9^{6-x}$
$3^{-4x}=3^{2(6-x)}$
$-4x=12-2x$
$2x=-12$
$x=-6$

worked example: solve $2^t=7+8\cdot 2^{-t}$
$2^t=7+2^3\cdot 2^{-t}$
since we have the number 7 being a pain, try to form a quadratic with substitution
$2^t=7+2^3\cdot \frac{1}{2^t}$
$2^t=7+\frac{8}{2^t}$
let $k=2^t$
$k=7+\frac{8}{k}$
$k^2=7k+8$
$(k+1)(k-8)=0$
$k=-1\vee k=8$
$2^t=-1\vee 2^t=8$
$2^t=-1$ is rejected because $2^a>0$
$t=3$

worked example: solve $3^{2x}+3=4\cdot 3^x$
try to form a quadratic with substitution
$(3^x{)}^2+3=4\cdot 3^x$
let $k=3^x$
$k^2-4k+3=0$
$(k-3)(k-1)=0$
$3^x=3\vee 3^x=1$
$x=1\vee x=0$

worked example: find solutions to $3^{2x}-4\cdot 3^x>-3$
rearrange the equation into a quadratic
let $k=3^x$
$k^2-4k+3>0$
$(k-3)(k-1)>0$
$k-3>0\vee k-1>0$
$k>3\vee k>1$
$3^x>3\vee 3^x>1$
$x>{\log }_3(3)\vee x>{\log }_3(1)$
$x>1\vee x>0$
simplify
$x>0$

worked example: for $f(x)=2^{1-x}$, simplify $\frac{f(x)}{f(-x)}$ into the form $2^{ax}$
$\frac{f(x)}{f(-x)}$
$\frac{2^{1-x}}{2^{1+x}}$
$2^{1-x-1-x}$
$2^{-2x}$

worked example: solve $5^{2x+1}>5^x+4$
try to form a quadratic with substitution
$5^1\cdot 5^{2x}>5^x+4$
let $k=5^x$
$5k^2>k+4$
$5k^2-k-4>0$
$5k^2-5k+4k-4>0$
$5k(k-1)+-4(k-1)>0$
$5k-4>0\vee k-1>0$
$k>\frac{4}{5}\vee k>1$
$5^x>\frac{4}{5}\vee 5^x>1$
$x>{\log }_5(\frac{4}{5})\vee x>0$

worked example: solve $\frac{21}{2^{2x+3}}\le 2^{2x+1}+2$
try to form a quadratic with substitution
$\frac{21}{8\cdot 2^{2x}}\le 2\cdot 2^{2x}+2$
let $k=2^x$
$\frac{21}{8k^2}\le 2k^2+2$
note that we need to keep the exponents of the substitution variable at either 1 or 2 to form a quadratic
let $m=2^{2x}$
$\frac{21}{8m}\le 2m+2$
$21\le 16m^2+16m$
$0\le 16m^2+16m-21$
this is hard to factorize but try anyways shrug
$0\le (4m+7)(4m-3)$
$4m+7\ge 0\vee 4m-3\ge 0$
$m\ge -\frac{7}{4}\vee m\ge \frac{3}{4}$
$2^{2x}\ge -\frac{7}{4}\vee 2^{2x}\ge \frac{3}{4}$
$2^{2x}\ge -\frac{7}{4}$ is rejected because $2^a>0$
$2x\ge {\log }_2(\frac{3}{4})$
$x\ge \frac{1}{2}{\log }_2(\frac{3}{4})$

worked example: solve $4\cdot 3^{x+1}-72=3^{x+2}+3^{x-1}$
try to form a quadratic with substitution
$4\cdot 3\cdot 3^x-72=9\cdot 3^x+3^{-1}\cdot 3^x$
let $k=3^x$
$12k-72=9k+\frac{k}{3}$
since the exponents of $k$ are all the same, quadratic does not need to be formed
$3k-72=\frac{k}{3}$
$8k=216$
$k=27$
$x=3$

worked example: solve $3\cdot 2^{\frac{x}{2}}-20=7\cdot 2^{\frac{x}{4}}$
try to form a quadratic with substitution
$3\cdot (2^x{)}^{\frac{1}{2}}-20=7\cdot (2^x{)}^{\frac{1}{4}}$
let $m=2^{\frac{1}{2}x}$
$3\cdot 2^{\frac{1}{2}x}-20=7\cdot (2^{\frac{1}{2}x}{)}^{\frac{1}{2}}$
$3m-20=7\sqrt{m}$
note that we need to keep the exponents of the substitution variable at either 1 or 2 to form a quadratic
let $k=2^{\frac{1}{4}x}$
$3\cdot (2^{\frac{1}{4}x}{)}^2-20=7\cdot 2^{\frac{1}{4}x}$
$3k^2-20=7k$
$3k^2-7k-20=0$
$3k^2-12k+5k-20=0$
$3k(k-4)+5(k-4)=0$
$k=-\frac{5}{3}\vee k=4$
$2^{\frac{1}{4}x}=-\frac{5}{3}\vee 2^{\frac{1}{4}x}=4$
$2^{\frac{1}{4}x}=-\frac{5}{3}$ is rejected because $2^a>0$
$2^{\frac{1}{4}x}=4$
$\frac{1}{4}x={\log }_2(4)$
$\frac{1}{4}x=2$
$x=8$

worked example: solve $3\cdot 9^{x+1}+1=25\cdot 3^x+3^{x+1}$
exponents are 1 away from each other, could use quadratics
$27\cdot 9^x+1=25\cdot 3^x+3\cdot 3^x$
$27\cdot (3^x{)}^2+1=25\cdot 3^x+3\cdot 3^x$
let $k=3^x$
$27k^2+1=28k$
$27k^2-28k+1=0$
$27k^2-27k-k+1=0$
$27k(k-1)+-1(k-1)=0$
$27k-1=0\vee k-1=0$
$k=\frac{1}{27}\vee k=1$
$3^x=\frac{1}{27}\vee 3^x=1$
$x=-3\vee x=0$

worked example: solve $25^x-\frac{1}{5^{1-2x}}=48$
$25^x-(5^{2x-1})=48$
$5^{2x}-(\frac{1}{5}\cdot 5^{2x})=48$
$(5^x{)}^2-(\frac{1}{5}\cdot (5^x{)}^2=48$
let $A=5^x$
$A^2-\frac{1}{5}{A}^2=48$
$\frac{4}{5}{A}^2=48$
$A^2=60$
$A=\pm \sqrt{60}$
$5^x=\pm \sqrt{60}$
$x=\pm {\log }_5(\sqrt{60})$

worked example: solve $\frac{5}{{\log }_e(x)+2}={\log }_e(x)-2$
let $A={\log }_e(x)+2$
$\frac{5}{A+2}=A-2$
$5=A^2-4$
$A^2=9$
$A=\pm 3$

worked example: solve $4^x+200\cdot 25^x=30\cdot 10^x$
try to make the exponent bases the same
this can be difficult but try to make the bases contain a small variety of numbers
$(2^x{)}^2+200\cdot (5^x{)}^2=30\cdot (2\cdot 5{)}^x$
$(2^x{)}^2+200\cdot (5^x{)}^2=30\cdot 2^x\cdot 5^x$
try to combine bases and exponents
think of a few ways… the easiest way to make this equation simpler is to divide both sides by $(5^x{)}^2$
$\frac{(2^x{)}^2}{(5^x{)}^2}+200=30\cdot 2^x\cdot \frac{5^x}{(5^x{)}^2}$
$\frac{(2^x{)}^2}{(5^x{)}^2}+200=30\cdot \frac{2^x}{5^x}$
now we can finally make all bases the same
${\left({\left(\frac{2}{5}\right)}^x\right)}^2+200=30\cdot {\left(\frac{2}{5}\right)}^x$
let $A={\left(\frac{2}{5}\right)}^x$
$A^2-30A+200=0$
$(A-20)(A-10)=0$
$A=20\vee A=10$
${\left(\frac{2}{5}\right)}^x=20\vee {\left(\frac{2}{5}\right)}^x=10$
$x={\log }_{\frac{2}{5}}(20)\vee x={\log }_{\frac{2}{5}}(10)$

worked example: solve $25^{4x+2}=125$
$5^{8x+4}=5^3$
$8x+4=3$
$x=-\frac{1}{8}$

solve $2^x=8-4\cdot 2^{-x+2}$

$2^x=8-4\cdot 4\cdot 2^{-x}$
$2^x=8-16\cdot \frac{1}{2^x}$
let $A=2^x$
$(2^x{)}^2=8\cdot 2^x-16$
$A^2=8A-16$
$A^2-8A+16=0$
$(A-4)(A-4)=0$
$A=4$
$2^x=4$
$x=2$