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

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

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

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

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

Solution:
[tex]\left(2^x+2^{-x}\right)^2=4^x+2+4^{-x}=25[/tex]
[tex]2^x+2^{-x}\gt 0 \Rightarrow 2^x+2^{-x}=5[/tex]

Solution:
[tex](-2)^{-2}=\dfrac{1}{4}[/tex]
[tex]8^{-\frac{5}{6}}\times 4^{2. 5}\times 2^{0. 5}=2^{-\frac{5}{2}}\times 2^5\times 2^{\frac{1}{2}}=2^3=8[/tex]
[tex]\dfrac{1}{4}+8=\dfrac{33}{4}[/tex]

Solution:
[tex]2+2^2+\ldots+2^n=2^{n+1}-2 \Rightarrow 2+2^2+\ldots+2^{2006}=2^{2007}-2[/tex]
[tex]2^{2007}-\left(2^{2007}-2\right)=2[/tex]

Solution:
[tex]0. 2=5^{-1}, 0. 04=5^{-2}, 125=5^3[/tex]
[tex]\dfrac{5^n\times 5^{1-n}}{5^{6-6n}\times 5^4}=\dfrac{5}{5^{10-6n}}=5^{6n-9}[/tex]

Solution:
[tex]\left(-2^3\right)^2=2^6, (-16)^3=-2^{12}, (-256)^{-3}=-2^{-24}, -(-512)^{-2}=-2^{-18}[/tex]
[tex]\dfrac{2^6\times \left(-2^{12}\right)\times \left(-2^{-24}\right)}{-2^{-18}}=\dfrac{2^{-6}}{-2^{-18}}=-2^{12}[/tex]

Solution:
The exponents [tex]-5, 0, 5, \ldots, 80[/tex] form an arithmetic progression with difference [tex]5[/tex] and [tex]18[/tex] terms.
[tex]n=\dfrac{(-5+80)\times 18}{2}=675[/tex]

Solution:
[tex]2^{2400}=16^{600}, 3^{3600}=729^{600}, 5^{1200}=25^{600}[/tex]
[tex]16\lt 25\lt 729 \Rightarrow 2^{2400}\lt 5^{1200}\lt 3^{3600}[/tex]

Solution:
[tex]3^{1333}=3\times \left(3^2\right)^{666}=3\times 9^{666}, 2^{1999}=2\times \left(2^3\right)^{666}=2\times 8^{666}[/tex]
[tex]3\gt 2, 9^{666}\gt 8^{666} \Rightarrow 3^{1333}\gt 2^{1999}[/tex]

Solution:
[tex]p=x^{\frac{1}{3}}, q=y^{\frac{1}{3}}:\ x-y=p^3-q^3=(p-q)\left(p^2+pq+q^2\right)[/tex]
[tex]\dfrac{p^2+q^2}{x-y}-\dfrac{p^2+pq+q^2}{x-y}=-\dfrac{pq}{x-y}=-\dfrac{\sqrt[3]{xy}}{x-y}[/tex]
[tex]\sqrt[3]{16\times 108}=\sqrt[3]{1728}=12, x-y=-92 \Rightarrow -\dfrac{12}{-92}=\dfrac{3}{23}[/tex]