Exponentiation problems: Difficult Problems with Solutions

Solution:
[tex]3^{-4} = \frac{1}{3^4} = \frac{1}{81}[/tex]

Solution:
[tex]\left(x^{\frac{1}{3}}\right)^3 = x^{\frac{1}{3}\cdot 3} = x[/tex]

Solution:
[tex]15-[16\times(-2)^{-3}] = 15 - (16\times\frac{1}{-2^3}) = 15 - [16\times(-\frac{1}{2^3})] = \\ = 15 - [16\times(-\frac{1}{8})] = 15 - (-2) = 15 + 2 = 17[/tex]

Solution:
We multiply on the right side: [tex]3^6 \times 5^2 \times 5^3 \times 3^{-1}=3^{6-1} \times 5^{3+2}=3^5 \times 5^5=15^5=15^n[/tex], so [tex]n=5[/tex].

Solution:
We multiply on the right side and get [tex]3^8 \times 6^3 \times 4^{10} \times 2^{-1}=3^8 \times 2^3 \times 3^3 \times 2^{20} \times 2^{-1}=3^{8+3} \times 2^{3+20-1}=3^{11} \times 2^{22}=3^{11} \times 4^{11}=12^{11}=12^n[/tex], so [tex]n=11[/tex].

Solution:
Since [tex]5^{2n+1}=5^7[/tex], [tex]2n+1=7[/tex]. The linear equation has only one solution and it is [tex]n=3[/tex].

Solution:
[tex]\frac{2^3 \times 3^2 \times 6^8}{2^{10} \times 3^6}=2^3 \times 3^2 \times 2^8 \times 3^8 \times 2^{-10} \times 3^{-6} = 2^{3+8-10} \times 3^{10-6}=2 \times 3^4=2 \times 81=162[/tex]

Solution:
[tex]\frac{3^8 \times 8^{11} \times 12^2}{27^3 \times 16^8}=3^8 \times (2^3)^{11} \times 3^2 \times 4^2 \times 27^{-3} \times 16^{-8} = 3^8 \times 2^{33} \times 3^2 \times 2^4 \times (3^3)^{-3} \times (2^4)^{-8} = 3^8 \times 2^{33} \times 3^2 \times 2^4 \times 3^{-9} \times 2^{-32} = 3^{8+2-9} \times 2^{33+4-32}=3^1 \times 2^5=3 \times 32=96[/tex]

Solution:
[tex]\frac{2^n}{4}=\frac{2^n}{2^2}=2^{n-2}=2^5, so n-2=5 or n=7.[/tex]

Solution:
[tex]\frac{7^3 \times 2^4 \times 3^5}{14^2 \times 21 \times 27}=7^3 \times 2^4 \times 3^5 \times 14^{-2} \times 21^{-1} \times 27^{-1} = 7^3 \times 2^4 \times 3^5 \times 2^{-2} \times 7^{-2} \times 3^{-1} \times 7^{-1} \times (3^3)^{-1} = 7^3 \times 2^4 \times 3^5 \times 2^{-2} \times 7^{-2} \times 3^{-1} \times 7^{-1} \times 3^{-3} = 7^{3-2-1} \times 2^{4-2} \times 3^{5-1-3}=7^0 \times 2^2 \times 3^1=1 \times 4 \times 3=12[/tex]

Solution:
[tex]2\times (2x)^3\times (8x)^2=2\times 2^3x^3\times 2^6x^2=2^{10}x^5[/tex]
[tex]2^{10}x^5=(2^2)^5x^5=(4x)^5[/tex]

Solution:
[tex]3^{k+2}=3^k\times 3^2=9\times 3^k[/tex]
[tex]3^k+3^k+3^k=3\times 3^k[/tex]
[tex]\frac{9\times 3^k}{3\times 3^k}=3[/tex]

Solution:
An odd power of a negative number is negative, an even power is positive:
[tex](-2)^{2007}=-2^{2007}[/tex], [tex](-2)^{2008}=2^{2008}[/tex]
[tex]x=\frac{-2^{2007}\times 2^{2008}}{2^{2007}\times 2^{2008}}=-1[/tex]
[tex]-x^2-5x+8=-(-1)^2-5\times (-1)+8=-1+5+8=12[/tex]

Solution:
Let's look at the last digits of the powers of [tex]9[/tex]:
[tex]9^1=9[/tex], [tex]9^2=81[/tex], [tex]9^3=729[/tex], [tex]9^4=6561[/tex], [tex]\ldots[/tex]
Odd powers of [tex]9[/tex] end in [tex]9[/tex], even powers end in [tex]1[/tex]. Since [tex]99[/tex] is odd, [tex]9^{99}[/tex] ends in [tex]9[/tex], and [tex]1+9^{99}[/tex] ends in [tex]0[/tex].

Solution:
Be careful: [tex]-2^2=-4[/tex] (the minus is not raised to the power), so [tex](-2^2)^3=(-4)^3=-64=-2^6[/tex].
[tex](-27)^2=3^6[/tex], [tex](-2)^5=-2^5[/tex], [tex](-9)^2=3^4[/tex]
[tex]\frac{-2^6\times 3^6}{-2^5\times 3^4}=2\times 3^2=18[/tex]

Solution:
[tex]4^{-2}=(2^2)^{-2}=2^{-4}[/tex], [tex]8^2=(2^3)^2=2^6[/tex]
[tex]\frac{2^5\times 2^{-4}}{2^6\times 2^{-3}}=\frac{2^{1}}{2^{3}}=2^{-2}=\frac{1}{4}[/tex]

Solution:
First we simplify:
[tex](-2x)^6=64x^6[/tex], [tex](-x)^5=-x^5[/tex], [tex](4x^3)^3=64x^9[/tex]
[tex]\frac{64x^6\times (-x^5)}{64x^9}=-\frac{x^{11}}{x^9}=-x^2[/tex]
[tex]x=-10[/tex]: [tex]-(-10)^2=-100[/tex]

Solution:
[tex]125^5=(5^3)^5=5^{15}[/tex]
[tex]5^{17}-2\times 5^{16}-3\times 5^{15}=5^{15}(5^2-2\times 5-3)=5^{15}(25-10-3)=12\times 5^{15}[/tex]
[tex]A=\frac{12\times 5^{15}}{5^{15}}=12[/tex]

Solution:
[tex]1. 1^5+1. 1^7=1. 1^5(1+1. 1^2)=1. 1^5(1+1. 21)=1. 1^5\times 2. 21[/tex]
[tex]\frac{1. 1^5\times 2. 21}{1. 1^5\times 2. 21}=1[/tex]

Solution:
[tex]49^0=1[/tex], [tex]\left(\frac{1}{7}\right)^{-2}=7^2=49[/tex]
[tex]1\frac{1}{7}\times 49=\frac{8}{7}\times 49=56[/tex]
[tex]8^{-8}\times 0. 125^{-8}=(8\times 0. 125)^{-8}=1^{-8}=1[/tex]
[tex]56-1=55[/tex]

Solution:
[tex]7^{7^7}[/tex] means [tex]7^{(7^7)}[/tex], and [tex]7^7=823\,543>77[/tex], so [tex]7^{7^7}>7^{77}[/tex].
[tex]77\lt 343=7^3[/tex], so [tex]77^7\lt (7^3)^7=7^{21}\lt 7^{77}[/tex].
[tex]777\lt 7^4=2401\lt 7^{77}[/tex]
The largest is [tex]7^{7^7}[/tex].