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: $0$