Exponential Equations: Problems with Solutions

Solution:
$3^{2x+1}=27$
$3^{2x+1}=3^3$
$2x+1=3$
$2x=2$
$x=1$

Solution:
$\sqrt[3]{2^x}=128$
$2^{\frac{x}{3}}=2^7$
$\frac{x}{3}=7$
$x=21$

Solution:
$\left(\frac{1}{3}\right)^{x^2-4x+4.5}=\sqrt{\frac{1}{3}}$

$\left(\frac{1}{3}\right)^{x^2-4x+4.5}=\left(\frac{1}{3}\right)^{\frac{1}{2}}$

$x^2-4x+4.5=\frac{1}{2}$

$x^2-4x+4.5=0.5$

$x^2-4x+4.5-0.5=0$

$x^2-4x+4=0$

$(x-2)^2=0$

$x=2$

Solution:
$\left(\frac56\right)^x=\left(\frac{36}{25}\right)^2$

$\left(\frac56\right)^x=\left(\frac{25}{36}\right)^{-2}$

$\left(\frac56\right)^x=\left(\frac{5^2}{6^2}\right)^{-2}$

$\left(\frac56\right)^x=\left(\left(\frac{5}{6}\right)^2\right)^{-2}$

$\left(\frac56\right)^x=\left(\frac{5}{6}\right)^{-4}$

$x=-4$

Solution:
Domain: $x+2\ne0 \Leftrightarrow x\ne2$

$\left(\frac{1}{2}\right)^{\frac{x-1}{x+2}}=4$

$\left(2^{-1}\right)^{\frac{x-1}{x+2}}=2^2$

$-\frac{x-1}{x+2}=2$

$\frac{1-x}{x+2}=2$

$1-x=2(x+2)$

$1-x=2x+4$

$-3x=3$

$x=-1$

Solution:
Domain: $4^{3x-2} \ge 0 \Leftrightarrow \forall x$

$\sqrt[3]{8^{x+2}}=\sqrt{4^{3x-2}}$

$8^{\frac{x+2}{3}}=4^{\frac{3x-2}{2}}$

$\left(2^{3}\right)^{\frac{x+2}{3}}=\left(2^{2}\right)^{\frac{3x-2}{2}}$

$2^{\frac{3(x+2)}{3}}=2^{\frac{2(3x-2)}{2}}$

$2^{x+2}=2^{3x-2}$

$x+2=3x-2$

$-2x=-4$

$2x=4$

$x=2$

Answer: $2$

Solution:
$6^{x^2+5}=\frac{1}{64^x\cdot3^{6x}}$

Domain: $64^x\cdot3^{6x} \ne 0 \Leftrightarrow \forall x$

$6^{x^2+5}=\frac{1}{(2^6)^x\cdot3^{6x}}$

$6^{x^2+5}=\frac{1}{2^{6x}\cdot3^{6x}}$

$6^{x^2+5}=\frac{1}{(2\cdot3)^{6x}}$

$6^{x^2+5}=\frac{1}{6^{6x}}$

$6^{x^2+5}=\left(\frac{1}{6}\right)^{6x}$

$6^{x^2+5}=6^{-6x}$

$x^2+5=-6x$
$x^2+6x+5=0$
$D=16$
$x_1=-5$
$x_2=-1$

Answer: $-5, -1$

Solution:
$9+9^x=10\cdot 3^x$

$9+\left(3^2\right)^x=10\cdot 3^x$

$9+3^{2x}=10\cdot 3^x$

$9+\left(3^x\right)^2=10\cdot 3^x$

Let $3^x=y, \ y > 0$

$9+y^2=10y$
$y^2-10y+9=0$
$D=64$
$y_1=9>0$
$y_2=1>0$

$3^x=9 \Leftrightarrow 3^x=3^2 \Rightarrow x=2$
$3^x=1 \Leftrightarrow 3^x=3^0 \Rightarrow x=0$

Answer: $0, 2$

Solution:
$9^x+2\cdot3^x-15=0$

$(3^2)^x+2\cdot3^x-15=0$

Let $3^x=y, \ y>0$

$y^2+2y-15=0$
$D=64$
$y_1=3 > 0$
$y_2=-5 < 0$ but it must be a positive number.
$3^x=3 \Rightarrow x=1$

Answer: $1$

Solution:
$3\cdot25^x-2\cdot5^x-1=0$

$3\cdot(5^2)^x-2\cdot5^x-1=0$

$3\cdot(5^x)^2-2\cdot5^x-1=0$

Let $5^x=y, \ y > 0$

$3y^2-2y-1=0$
$D=16$
$y_1=1 > 0$
$y_2=-\frac{1}{3}<0$ - impossible

$3^x=1 \Leftrightarrow 3^x=3^0\Rightarrow x = 0$

Answer: $0$

Solution:
$2\cdot3^{3x-1}+27^{x-\frac{2}{3}}=9^{x-1}+2\cdot3^{2x-1}$

$2\cdot\frac{3^{3x}}{3}+\frac{27^x}{27^{\frac{2}{3}}}=\frac{9^x}{9}+2\cdot\frac{3^{2x}}{3}$

$2\cdot\frac{(3^{x})^3}{3}+\frac{(3^x)^3}{(\sqrt[3]{27})^2}=\frac{(3^x)^2}{9}+2\cdot\frac{(3^x)^2}{3}$

$2\cdot\frac{(3^{x})^3}{3}+\frac{(3^x)^3}{9}=\frac{(3^x)^2}{9}+2\cdot\frac{(3^x)^2}{3}$

Let $3^{x} = y,\ \ y > 0$

$2\cdot\frac{y^3}{3}+\frac{y^3}{9}=\frac{y^2}{9}+2\cdot\frac{y^2}{3}$

$9\cdot2\cdot\frac{y^3}{3}+9\cdot\frac{y^3}{9}=9\cdot\frac{y^2}{9}+9\cdot2\cdot\frac{y^2}{3}$

$6y^3+y^3=y^2+6y^2$

$7y^3=7y^2$

$y^3=y^2$

$y^3-y^2=0$

$y^2(y-1)=0$

$y_1=0$ but $y > 0$ - so we cannot continue with this value
$y_2=1>0$

$3^x=1 \Leftrightarrow 3^x=3^0 \Rightarrow x=0$

Answer: $0$

Solution:
[tex]1=3^0 \Rightarrow (x^2+x-2)(3-x)=0 \iff (x+2)(x-1)(3-x)=0[/tex]
[tex]x_1=-2, x_2=1, x_3=3[/tex]

Solution:
Substitute [tex]y=x^2-3x+3[/tex]. Dividing by [tex]7\times 8[/tex]:
[tex]7^{y-1}=8^{y-1} \iff \left(\dfrac{7}{8}\right)^{y-1}=1 \Rightarrow y=1[/tex]
[tex]x^2-3x+3=1 \iff x^2-3x+2=0 \Rightarrow x_1=1, x_2=2[/tex]

Solution:
[tex]3^{2x-1-3x}\times 5^{3x+2-2x}=3^2\times 5^{-1}[/tex]
[tex]3^{-x-1}\times 5^{x+2}=3^2\times 5^{-1} \iff 5^{x+3}=3^{x+3} \iff \left(\dfrac{5}{3}\right)^{x+3}=1[/tex]
[tex]x+3=0 \Rightarrow x=-3[/tex]

Solution:
[tex]4^{x+y}=2^{2x+2y}=2^{10} \Rightarrow x+y=5[/tex]
[tex]2^{5x-2y}=2^4 \Rightarrow 5x-2y=4[/tex]
[tex]y=5-x:\ 5x-10+2x=4 \Rightarrow x=2, y=3[/tex]

Solution:
[tex]5^{x-1}\times 2^{x-3}=\dfrac{5^x\times 2^x}{5\times 8}=\dfrac{10^x}{40}[/tex]
[tex]10^x-\dfrac{10^x}{40}=\dfrac{39}{40}\times 10^x=975 \Rightarrow 10^x=1000 \Rightarrow x=3[/tex]

Solution:
[tex]5^{|4x-6|}=5^{6x-8} \iff |4x-6|=6x-8, 6x-8\ge 0 \iff x\ge \dfrac{4}{3}[/tex]
Case [tex]x\ge \dfrac{3}{2}:\ 4x-6=6x-8 \Rightarrow x=1\lt \dfrac{3}{2}[/tex] — is rejected.
Case [tex]x\lt \dfrac{3}{2}:\ 6-4x=6x-8 \Rightarrow x=\dfrac{7}{5}\in\left[\dfrac{4}{3}, \dfrac{3}{2}\right)\ \checkmark[/tex]

Solution:
[tex]f(x)=2^x+x-3[/tex] is strictly increasing (a sum of increasing functions), so there is at most one root.
[tex]f(1)=2+1-3=0[/tex]
Answer: [tex]x=1[/tex]

Solution:
[tex]\dfrac{3^x}{9}=5^x \iff \left(\dfrac{3}{5}\right)^x=9 \Rightarrow x=\log_{\frac{3}{5}}9[/tex]

Solution:
[tex]\dfrac{4^x}{4}\times 5\times 25^x=\dfrac{5}{4}\times 100^x=80 \Rightarrow 100^x=64[/tex]
[tex]10^{2x}=8^2 \Rightarrow 10^x=8 \Rightarrow x=\log 8[/tex]

Solution:
Substitute [tex]t=5^x\gt 0, t\ne 3[/tex]: [tex]8(t+1)-6(t-3)=3(t-3)(t+1)[/tex]
[tex]2t+26=3t^2-6t-9 \iff 3t^2-8t-35=0 \Rightarrow t_1=5, t_2=-\dfrac{7}{3}[/tex]; [tex]t_2[/tex] is rejected.
[tex]5^x=5 \Rightarrow x=1[/tex]

Solution:
The sum of the first [tex]x[/tex] odd numbers is [tex]x^2[/tex], so the left side is [tex]2^{1+3+5+\ldots+(2x-1)}=2^{x^2}[/tex].
[tex]0. 25^{-18}=4^{18}=2^{36}[/tex]
[tex]x^2=36, x\in\mathbb{N} \Rightarrow x=6[/tex]