Exponential Equations: Very Difficult Problems with Solutions

Solution:
[tex]\log_{\frac{1}{27}}\dfrac{1}{3}=\dfrac{1}{3} \Rightarrow (x+y)\times 2^{y-x}=3[/tex]
Substitute [tex]s=x+y\gt 0, d=x-y\ne 0[/tex]: [tex]s=3\times 2^d[/tex]
[tex]\left(3\times 2^d\right)^{\frac{1}{d}}=2\times 3^{\frac{1}{d}}=2\sqrt{3} \Rightarrow 3^{\frac{1}{d}}=3^{\frac{1}{2}} \Rightarrow d=2, s=12[/tex]
[tex]x+y=12, x-y=2 \Rightarrow x=7, y=5[/tex]

Solution:
[tex]7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2, 26+15\sqrt{3}=\left(2+\sqrt{3}\right)^3, \left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=1[/tex]
Substitute [tex]t=\left(2+\sqrt{3}\right)^x\gt 0[/tex]: [tex]t^3-5t^2+6t+\dfrac{1}{t}=5 \iff t^4-5t^3+6t^2-5t+1=0[/tex]
Dividing by [tex]t^2[/tex], [tex]s=t+\dfrac{1}{t}\ge 2[/tex]: [tex]t^2+\dfrac{1}{t^2}-5\left(t+\dfrac{1}{t}\right)+6=0 \iff s^2-5s+4=0[/tex]
[tex]s=4[/tex]; [tex]s=1[/tex] is rejected.
[tex]t^2-4t+1=0 \Rightarrow t=2\pm\sqrt{3}=\left(2+\sqrt{3}\right)^{\pm 1} \Rightarrow x=\pm 1[/tex]

Solution:
Substitute [tex]t=2^x\gt 0[/tex]: [tex]3t^2+(3x-10)t+(3-x)=0[/tex]
[tex]D=(3x-10)^2-12(3-x)=9x^2-48x+64=(3x-8)^2[/tex]
[tex]t=\dfrac{10-3x\pm(3x-8)}{6}:\ t=\dfrac{1}{3}, t=3-x[/tex]
[tex]2^x=\dfrac{1}{3} \Rightarrow x=-\log_2 3[/tex]
[tex]2^x=3-x[/tex]: The left side is strictly increasing and the right side is strictly decreasing, so the equation has at most one root. [tex]x=1:\ 2=2\ \checkmark[/tex]

Solution:
Substitute [tex]t=2^x\gt 0[/tex]: [tex]t^2-(a+3)t+4(a-1)=0, D=(a+3)^2-16(a-1)=(a-5)^2[/tex]
[tex](t-4)(t-a+1)=0[/tex]
The root [tex]t=4[/tex], i.e. [tex]x=2[/tex], always exists. The second root [tex]t=a-1[/tex] gives another [tex]x[/tex] exactly when [tex]a-1\gt 0[/tex] and [tex]a-1\ne 4[/tex]. So the root is unique when [tex]a\le 1[/tex] or [tex]a=5[/tex].
Answer: [tex]a\in(-\infty, 1]\cup\{5\}[/tex]

Solution:
Case [tex]x\ge -1:\ 2^{x+2}-2^{x+1}+1=2^{x+1}+1[/tex]
This is an identity ([tex]2^{x+2}=2\times 2^{x+1}[/tex]), so every [tex]x\ge -1[/tex] is a solution.
Case [tex]-2\le x\lt -1:\ 2^{x+2}-1+2^{x+1}=2^{x+1}+1 \Rightarrow 2^{x+2}=2 \Rightarrow x=-1\notin[-2, -1)[/tex]
Case [tex]x\lt -2:\ 2^{-x-2}-1+2^{x+1}=2^{x+1}+1 \Rightarrow 2^{-x-2}=2 \Rightarrow x=-3\ \checkmark[/tex]
Answer: [tex]x\in\{-3\}\cup[-1, +\infty)[/tex]

Solution:
Substitute [tex]a=3^x, b=2^x, c=1[/tex]: [tex]a^2+b^2+c^2=ab+bc+ca[/tex]
[tex]a^2+b^2+c^2-ab-bc-ca=\dfrac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]=0[/tex]
[tex]a=b=c \Rightarrow 3^x=2^x=1 \Rightarrow x=0[/tex]

Solution:
Domain: [tex]6^x\ne 3^x \iff x\ne 0[/tex].
[tex]\dfrac{2^x\left(2^x-1\right)\left(2^x+1\right)}{3^x\left(2^x-1\right)}=2 \iff 4^x+2^x=2\times 3^x \iff \left(\dfrac{4}{3}\right)^x+\left(\dfrac{2}{3}\right)^x=2[/tex]
The function [tex]g(x)=\left(\dfrac{4}{3}\right)^x+\left(\dfrac{2}{3}\right)^x[/tex] is strictly convex (a sum of exponential functions), so it takes the value [tex]2[/tex] at most twice; [tex]g(0)=g(1)=2[/tex]. Since [tex]x\ne 0[/tex], [tex]x=1[/tex].

Solution:
By the AM–GM inequality:
[tex]2^{x^6}+2^{x^2}\ge 2\sqrt{2^{x^6+x^2}}=2^{1+\frac{x^6+x^2}{2}}\ge 2^{1+x^4}[/tex]
[tex]x^6+x^2-2x^4=x^2\left(x^2-1\right)^2\ge 0[/tex]
Equality requires [tex]x^6=x^2[/tex] and [tex]x^2\left(x^2-1\right)^2=0 \Rightarrow x\in\{-1, 0, 1\}[/tex]

Solution:
[tex]\dfrac{\sqrt{2}}{2}=\left(\dfrac{1}{2}\right)^{\frac{1}{2}}=\left(\dfrac{1}{4}\right)^{\frac{1}{4}} \Rightarrow x=\dfrac{1}{2}, x=\dfrac{1}{4}[/tex]
[tex]g(x)=x\ln x=-\dfrac{\ln 2}{2}, g'(x)=\ln x+1[/tex]
[tex]g[/tex] decreases on [tex]\left(0, \dfrac{1}{e}\right][/tex] and increases on [tex]\left[\dfrac{1}{e}, +\infty\right)[/tex], so it takes each value at most twice. There are no other roots.