Exponential Equations: Difficult Problems with Solutions

Solution:
Dividing by [tex]81^x\gt 0[/tex]. Substitute [tex]t=\left(\dfrac{4}{9}\right)^x\gt 0[/tex], [tex]\left(\dfrac{16}{81}\right)^x=t^2, \left(\dfrac{36}{81}\right)^x=t[/tex]:
[tex]3t^2-5t+2=0 \Rightarrow t_1=1, t_2=\dfrac{2}{3}[/tex]
[tex]\left(\dfrac{4}{9}\right)^x=1 \Rightarrow x=0[/tex]
[tex]\left(\dfrac{4}{9}\right)^x=\dfrac{2}{3}=\left(\dfrac{4}{9}\right)^{\frac{1}{2}} \Rightarrow x=\dfrac{1}{2}[/tex]

Solution:
[tex]\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=16-15=1[/tex]
Substitute [tex]t=\left(4+\sqrt{15}\right)^x\gt 0[/tex]: [tex]t+\dfrac{1}{t}=62 \iff t^2-62t+1=0[/tex]
[tex]t=31\pm 8\sqrt{15}=\left(4\pm\sqrt{15}\right)^2=\left(4+\sqrt{15}\right)^{\pm 2} \Rightarrow x=\pm 2[/tex]

Solution:
Substitute [tex]t=3^x\gt 0[/tex]: [tex]t^3-13t^2+39t-27=0[/tex]
[tex](t^3-27)-13t(t-3)=(t-3)(t^2+3t+9-13t)=(t-3)(t^2-10t+9)=(t-3)(t-1)(t-9)[/tex]
[tex]t\in\{1, 3, 9\} \Rightarrow x\in\{0, 1, 2\}[/tex]

Solution:
Domain: [tex]x\ne -\dfrac{1}{2}[/tex].
[tex]\dfrac{2-x}{2x+1}=1-\dfrac{3x-1}{2x+1}[/tex]
Substitute [tex]t=2^{\frac{3x-1}{2x+1}}\gt 0[/tex]: [tex]t-1=\dfrac{2}{t} \iff t^2-t-2=0 \Rightarrow t=2[/tex]; [tex]t=-1[/tex] is rejected.
[tex]\dfrac{3x-1}{2x+1}=1 \Rightarrow 3x-1=2x+1 \Rightarrow x=2[/tex]

Solution:
[tex]x^2-2x-3=(x-3)(x+1)[/tex]. Taking logarithms to base [tex]7[/tex]:
[tex](x-3)\log_7 5=(x-3)(x+1) \iff (x-3)\left(x+1-\log_7 5\right)=0[/tex]
[tex]x_1=3, x_2=\log_7 5-1=\log_7\dfrac{5}{7}[/tex]

Solution:
[tex]6\times 35^x+10\times 21^x=2\times 7^x\left(3\times 5^x+5\times 3^x\right)[/tex]
[tex]9\times 5^x+5\times 3^{x+1}=3\left(3\times 5^x+5\times 3^x\right)[/tex]
The bracket is positive, so [tex]2\times 7^x=3 \Rightarrow x=\log_7\dfrac{3}{2}[/tex]

Solution:
Dividing by [tex]5^x[/tex]: [tex]\left(\dfrac{3}{5}\right)^x+\left(\dfrac{4}{5}\right)^x=1[/tex]
The left side is strictly decreasing, so the equation has at most one root.
[tex]x=2:\ \dfrac{9}{25}+\dfrac{16}{25}=1\ \checkmark[/tex]
Answer: [tex]x=2[/tex]

Solution:
Substitute [tex]t=3^x\gt 0[/tex]: [tex]|t-1|+|3-t|=2[/tex]
[tex]|t-1|+|3-t|\ge |(t-1)+(3-t)|=2[/tex], with equality exactly when [tex]1\le t\le 3[/tex].
[tex]1\le 3^x\le 3 \iff 0\le x\le 1[/tex]

Solution:
Domain: [tex]x\ge 1[/tex].
[tex]x=1:\ 1=1\ \checkmark[/tex]
[tex]x\gt 1[/tex]: the exponents are equal: [tex]\sqrt{x-1}=x-3, x\ge 3[/tex]
[tex]x-1=x^2-6x+9 \iff x^2-7x+10=0 \Rightarrow x=2\ (2\lt 3), x=5\ \checkmark[/tex]

Solution:
Substitute [tex]a=25^x\gt 0, b=25^y\gt 0[/tex]: [tex]a^2+b^2=30, ab=5\sqrt{5} \Rightarrow a^2b^2=125[/tex]
[tex]a^2, b^2[/tex] are the roots of [tex]z^2-30z+125=0 \Rightarrow z=25, z=5[/tex]
[tex]\{a, b\}=\{5, \sqrt{5}\}[/tex]; [tex]25^x=5 \Rightarrow x=\dfrac{1}{2}, 25^y=\sqrt{5} \Rightarrow y=\dfrac{1}{4}[/tex]
Answer: [tex](x, y)\in\left\{\left(\dfrac{1}{2}, \dfrac{1}{4}\right), \left(\dfrac{1}{4}, \dfrac{1}{2}\right)\right\}[/tex]

Solution:
The base [tex]\dfrac{1}{5}[/tex] is less than [tex]1[/tex], so
[tex]x\lt 0 \iff \left(\dfrac{1}{5}\right)^x\gt 1 \iff \dfrac{a+3}{5-a}\gt 1[/tex]
[tex]\dfrac{a+3}{5-a}-1=\dfrac{2a-2}{5-a}\gt 0 \iff 1\lt a\lt 5[/tex]

Solution:
Domain: [tex]|x|\ge\sqrt{2}[/tex].
Substitute [tex]t=2^{x+\sqrt{x^2-2}}\gt 0[/tex]: [tex]t^2-\dfrac{5}{2}t-6=0 \iff 2t^2-5t-12=0 \Rightarrow t=4[/tex]; [tex]t=-\dfrac{3}{2}[/tex] is rejected.
[tex]x+\sqrt{x^2-2}=2 \iff \sqrt{x^2-2}=2-x, x\le 2[/tex]
[tex]x^2-2=4-4x+x^2 \Rightarrow x=\dfrac{3}{2}\ \ \left(\dfrac{3}{2}\ge\sqrt{2}\right)[/tex]

Solution:
Substitute [tex]u=3^x+\dfrac{6}{3^x}\gt 0[/tex]: [tex]3^{2x}+\dfrac{36}{3^{2x}}=u^2-12[/tex]
[tex]u^2-6u-7=0 \Rightarrow u=7[/tex]; [tex]u=-1[/tex] is rejected.
[tex]t=3^x:\ t+\dfrac{6}{t}=7 \iff t^2-7t+6=0 \Rightarrow t=1, t=6[/tex]
[tex]x_1=0, x_2=\log_3 6[/tex]

Solution:
[tex]\left(1+\sqrt{3}\right)^2=4+2\sqrt{3}=2\left(2+\sqrt{3}\right) \Rightarrow 2^x\left(2+\sqrt{3}\right)^x=\left(1+\sqrt{3}\right)^{2x}[/tex]
Substitute [tex]t=\left(1+\sqrt{3}\right)^x\gt 0[/tex]: [tex]t+\dfrac{t^2}{2}=4 \iff t^2+2t-8=0 \Rightarrow t=2[/tex]; [tex]t=-4[/tex] is rejected.
[tex]\left(1+\sqrt{3}\right)^x=2 \Rightarrow x=\log_{1+\sqrt{3}}2[/tex]

Solution:
Domain: [tex]x\ne -1[/tex].
[tex]\dfrac{3^x}{9}=\dfrac{4}{8^{\frac{x}{x+1}}} \iff 3^{x-2}=2^{2-\frac{3x}{x+1}}=2^{\frac{2-x}{x+1}}[/tex]
Taking logarithms to base [tex]3[/tex]: [tex]x-2=-\dfrac{x-2}{x+1}\log_3 2 \iff (x-2)\left(1+\dfrac{\log_3 2}{x+1}\right)=0[/tex]
[tex]x=2[/tex] or [tex]x+1=-\log_3 2 \Rightarrow x=-1-\log_3 2=-\log_3 6[/tex]

Solution:
Substitute [tex]t=4^{x-1}\gt 0[/tex]: [tex]4^{x+\frac{1}{2}}=4^{x-1}\times 4^{\frac{3}{2}}=8t, 16^{x-1}=t^2[/tex]
[tex]16-8t+t^2=(t-4)^2, 17-8t+t^2=(t-4)^2+1[/tex]
[tex]2|t-4|=(t-4)^2+1 \iff \left(|t-4|-1\right)^2=0 \Rightarrow |t-4|=1[/tex]
[tex]t=3 \Rightarrow x=1+\log_4 3, t=5 \Rightarrow x=1+\log_4 5[/tex]