Exponents and Radicals: Difficult Problems with Solutions

Solution:
$2^{x+1}+2^{x}+2^{x-1}=28$
$2^{x+1}+2^{x}+2^{x-1}=4\cdot 7=2^{2}\cdot7$

$2^{x+1}+2^{x}+2^{x-1}=2^{x-1}(2^{2}+2^{1}+1)=2^{2} \cdot 7$
$2^{x-1}7=2^{2} \cdot 7$
$2^{x-1}=2^{2}$
$x-1=2$
$x=3$

Solution:
$4^{\sqrt{x+1}}-2^{\sqrt{x+1}+2}=0$
$(2^{2})^{\sqrt{x+1}}-2^{\sqrt{x+1}+2}=0$

$2^{2\sqrt{x+1}}-2^{\sqrt{x+1}+2}=0$
$2^{2\sqrt{x+1}}=2^{\sqrt{x+1}+2}$

$2\sqrt{x+1}=\sqrt{x+1}+2 $
$\sqrt{x+1}=2$
$x+1=4\Longrightarrow x=3$

Solution:
$5^{3x}-5^{3x+1}+5^{3x+2}=5^{3x}(1-5^{1}+5^{2})=3 \cdot 7 \cdot 25^{3}$

$5^{3x} \cdot 21=21 \cdot (5^{2})^{3}$
$5^{3x}=5^{6}$
$3x=6$
$x=2$

Solution:
$\frac{3}{3^{x}}-3^{x}=2$
$\frac{3-3^{2x}}{3^{x}}=2$
$3-3^{2x}=2\cdot 3^{x}$

$3-(3^{x})^{2}-2\cdot 3^{x}=0$ (multiply by -1)
Substitute $u=3^{x}$
$u^{2}+2u-3=0$
$(u-1) \cdot (u+3)=0$

$u=1$ or $u=-3$
Discard $u=-3$ ($3^x$ is impossible to be negative)
So $u=1$
$u=3^{x}=1=3^{0}$
$x = 0$

Solution:
$2-3^{-x}+3^{x+1}=2-\frac{1}{3^{x}}+3 \cdot 3^{x}=\frac{2\cdot 3^{x}-1+3 \cdot 3^{x} \cdot 3^{x}}{3^{x}}=0$

$3 \cdot 3^{2x}+2 \cdot 3^{x}-1=0$
$3\cdot (3^{x})^{2}+2 \cdot 3^{x}-1=0$

Substitute $u=3^{x}$
$3\cdot (u)^{2}+2 \cdot u-1=0$
$u=\frac{-2\pm \sqrt{4+12}}{6}=\frac{-2\pm 4}{6}$

$u_{1}=\frac{1}{3}\qquad u_{2}=-1$
We discard $u_{2}=-1$($3^x$ is always a positive number)

$u=3^{x}=\frac{1}{3}=3^{-1}\Longrightarrow x=-1$

Solution:
$56^{2x-8}=1=56^{0}$
$2x-8=0$
$x=4$

Solution:
$3^{7-2x}=(\frac{1}{27})^{-8}=(\frac{1}{3^{3}})^{-8}=(3^{-3})^{-8}=3^{24}$
$3^{7-2x}=3^{24}$
$7-2x=24$
$x=-\frac{17}{2}$

Solution:
$\left(\frac{9}{8}\right)^{3x+12}=1=\left(\frac{9}{8}\right)^{0}$

$3x+12=0$
$x=-4$

Solution:
$3^{7x+4}=\left(\frac{1}{27}\right)^{x-3}=\left(\frac{1}{3^{3}}\right)^{x-3}=\left(3^{-3}\right) ^{x-3}=3^{-3x+9}$

$3^{7x+4}=3^{-3x+9}$
$7x+4=-3x+9$
$10x=5$

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

Solution:
$2^{8x-4}\cdot 16^{3x-5}=16^{9x+2}$
$2^{8x-4}\cdot \left( 2^{4}\right) ^{3x-5}=\left( 2^{4}\right) ^{9x+2}$
$2^{8x-4}\cdot \left( 2\right) ^{12x-20}=\left( 2\right) ^{36x+8}$

$2^{8x-4+12x-20}=\left( 2\right) ^{36x+8}$
$8x-4+12x-20=36x+8$
$20x-24=36x+8$
$-32=16x$
$x=-2$

Solution:
$\left(\frac{1}{64}\right)^{-8x+3}=\left(\frac{1}{16}\right)^{7x-2}$

$\left(\frac{1}{2^{6}}\right)^{-8x+3}=\left(\frac{1}{2^{4}}\right)^{7x-2}$

$(2^{-6})^{-8x+3}=(2^{-4})^{7x-2}$
$2^{48x-18}=2^{-28x+8}$

$48x-18=-28x+8$
$76x=26$

$x=\frac{26}{76}=\frac{13}{38}$